Haskell, 37 bytes

f l=[x|x<-l,y<-l,zip x x==zip x y]==l

Each element x of l is repeated once for every element that it's a prefix of, which is exactly once for a prefix-free list, giving the original list. The prefix property is checked by zipping both lists with x, which cuts off elements beyond the length of x.

Java, 128 127 126 125 124 121 bytes

(Thanks @Kenny Lau, @Maltysen, @Patrick Roberts, @Joba)

Object a(String[]a){for(int i=0,j,l=a.length;i<l;i++)for(j=0;j<l;)if(i!=j&a[j++].startsWith(a[i]))return 1<0;return 1>0;}

Ungolfed

Object a(String[] a) {
    for (int i = 0, j, l = a.length; i < l; i++) 
        for (j = 0; j < l;) 
            if (i != j & a[j++].startsWith(a[i])) return 1<0;
    return 1>0;
}

Output

[Hello, World]
true

[Code, Golf, Is, Cool]
true

[1, 2, 3, 4, 5]
true

[This, test, case, is, true]
true

[111, 010, 000, 1101, 1010, 1000, 0111, 0010, 1011, 0110, 11001, 00110, 10011, 11000, 00111, 10010]
true

[4, 42]
false

[1, 2, 3, 34]
false

[This, test, case, is, false, t]
false

[He, said, Hello]
false

[0, 00, 00001]
false

[Duplicate, Duplicate, Keys, Keys]
false

Python 2, 48 51 bytes

lambda l:all(1/map(a.find,l).count(0)for a in l)

For each element a of l, the function a.find finds the index of the first occurrence of a in the input string, giving -1 for an absence. So, 0 indicates a prefix. In a prefix-free list, mapping this function returns only a single 0 for a itself. The function checks that this is the case for every a.

51 bytes:

lambda l:[a for a in l for b in l if b<=a<b+'~']==l

Replace ~ with a character with ASCII code 128 or higher.

For each element a in l, a copy is included for each element that's a prefix of it. For a prefix-free list, the only such element is a itself, so this gives the original list.

CJam, 14 bytes

q~$W%2ew::#0&!

Test suite.

Explanation

q~   e# Read and evaluate input.
$    e# Sort strings. If a prefix exists it will end up directly in front 
     e# of a string which contains it.
W%   e# Reverse list.
2ew  e# Get all consecutive pairs of strings.
::#  e# For each pair, find the first occurrence of the second string in the first.
     e# If a prefix exists that will result in a 0, otherwise in something non-zero.
0&   e# Set intersection with 0, yielding [0] for falsy cases and [] for truthy ones.
!    e# Logical NOT.

JavaScript ES6, 65 43 40 bytes

a=>!/(.*)\1/.test(''+a.sort().join``)
      ^            ^               ^ embedded NUL characters

My previous solution, which handled string arrays of all UTF-8 characters:

a=>!/[^\\]("([^"]*\\")*[^\\])",\1/.test(JSON.stringify(a.sort()))

I was able to avoid JSON.stringify since the challenge specifies printable ASCII characters only.

Test

f=a=>!/(\0.*)\1/.test('\0'+a.sort().join`\0`) // since stackexchange removes embedded NUL characters

O.textContent += 'OK: '+
[["Hello", "World"]                      
,["Code", "Golf", "Is", "Cool"]
,["1", "2", "3", "4", "5"]
,["This", "test", "case", "is", "true"]          
,["111", "010", "000", "1101", "1010", "1000", "0111", "0010", "1011", 
 "0110", "11001", "00110", "10011", "11000", "00111", "10010"]
].map(a=>f(a)) 

O.textContent += '\nKO: '+
[["4", "42"]                             
,["1", "2", "3", "34"]                   
,["This", "test", "case", "is", "false", "t"]
,["He", "said", "Hello"]
,["0", "00", "00001"]
,["Duplicate", "Duplicate", "Keys", "Keys"]
].map(a=>f(a))

<pre id=O></pre>

Haskell, 49 bytes

g x=[1|z<-map((and.).zipWith(==))x<*>x,z]==(1<$x)

This has a couple parts:

-- Are two lists (or strings) equal for their first min(length_of_1,length_of_2) elements, i.e. is one the prefix of the other?
(and.).zipWith(==)

-- Check whether one element is the prefix of the other, for all pairs of elements (including equal pairs)
map((and.).zipWith(==))x<*>x

-- This is a list of 1's of length (number of elements that are the prefix of the other)
[1|z<-map((and.).zipWith(==))x<*>x,z]

-- This is the input list, with all the elements replaced with 1's
(1<$x)

If the two lists are equal, then an element is only the prefix of itself, and it's valid.

Java, 97 bytes

Object a(String[]a){for(String t:a)for(String e:a)if(t!=e&t.startsWith(e))return 1<0;return 1>0;}

Uses most of the tricks found in @Marv's answer, but also makes use of the foreach loop and string reference equality.

Unminified:

Object a(String[]a){
    for (String t : a)
        for (String e : a)
            if (t != e & t.startsWith(e))
                return 1<0;
    return 1>0;
}

Note that, as I said, this uses string reference equality. That does mean that the code can behave oddly due to String interning. The code does work when using arguments passed from the command line, and also when using something read from the command line. If you want to hardcode the values to test, though, you'd need to manually call the String constructor to force interning to not occur:

System.out.println(a(new String[] {new String("Hello"), new String("World")}));
System.out.println(a(new String[] {new String("Code"), new String("Golf"), new String("Is"), new String("Cool")}));
System.out.println(a(new String[] {new String("1"), new String("2"), new String("3"), new String("4"), new String("5")}));
System.out.println(a(new String[] {new String("This"), new String("test"), new String("case"), new String("is"), new String("true")}));
System.out.println(a(new String[] {new String("111"), new String("010"), new String("000"), new String("1101"), new String("1010"), new String("1000"), new String("0111"), new String("0010"), new String("1011"), new String("0110"), new String("11001"), new String("00110"), new String("10011"), new String("11000"), new String("00111"), new String("10010")}));
System.out.println(a(new String[] {new String("4"), new String("42")}));
System.out.println(a(new String[] {new String("1"), new String("2"), new String("3"), new String("34")}));
System.out.println(a(new String[] {new String("This"), new String("test"), new String("case"), new String("is"), new String("false"), new String("t")}));
System.out.println(a(new String[] {new String("He"), new String("said"), new String("Hello")}));
System.out.println(a(new String[] {new String("0"), new String("00"), new String("00001")}));
System.out.println(a(new String[] {new String("Duplicate"), new String("Duplicate"), new String("Keys"), new String("Keys")}));

Retina, 19 bytes

Byte count assumes ISO 8859-1 encoding.

O`.+
Mm1`^(.+)¶\1
0

The input should be linefeed-separated. Output is 0 for falsy and 1 for truthy.

Try it online! (Slightly modified to support multiple space-separated test cases instead.)

Explanation

O`.+

Sort the lines in the input. If a prefix exists it will end up directly in front of a string which contains it.

Mm1`^(.+)¶\1

Try to match (M) a complete line which is also found at the beginning of the next line. The m activates multiline mode such that ^ matches line beginnings and the 1 ensures that we only count at most one match such that the output is 0 or 1.

0

To swap 0 and 1 in the result, we count the number of 0s.

Racket, 70 bytes

(λ(l)(andmap(λ(e)(not(ormap(curryr string-prefix? e)(remv e l))))l))

Python, 58 55 bytes

lambda l:sum(0==a.find(b)for a in l for b in l)==len(l)

JavaScript (ES6), 52 54

Edit 2 bytes saved thx @Neil

a=>!a.some((w,i)=>a.some((v,j)=>i-j&&!w.indexOf(v)))

Test

f=a=>!a.some((w,i)=>a.some((v,j)=>i-j&&!w.indexOf(v)))

O.textContent += 'OK: '+
[["Hello", "World"]                      
,["Code", "Golf", "Is", "Cool"]
,["1", "2", "3", "4", "5"]
,["This", "test", "case", "is", "true"]          
,["111", "010", "000", "1101", "1010", "1000", "0111", "0010", "1011", 
 "0110", "11001", "00110", "10011", "11000", "00111", "10010"]
].map(a=>f(a)) 

O.textContent += '\nKO: '+
[["4", "42"]                             
,["1", "2", "3", "34"]                   
,["This", "test", "case", "is", "false", "t"]
,["He", "said", "Hello"]
,["0", "00", "00001"]
,["Duplicate", "Duplicate", "Keys", "Keys"]
].map(a=>f(a))

<pre id=O></pre>

Mathematica 75 69 68 bytes

Loquacious as usual. But Martin B was able to reduce the code by 7 bytes.

Method 1: Storing output in an Array

(68 bytes)

f@a_:=!Or@@(Join@@Array[a~Drop~{#}~StringStartsQ~a[[#]]&,Length@a])

f@{"111", "010", "000", "1101", "1010", "1000", "0111", "0010", "1011", "0110", "11001", "00110", "10011", "11000", "00111", "10010"}

True

f@{"He", "said", "Hello"}

False

Method 2: Storing output in a List

(69 bytes)

f@a_:=!Or@@Flatten[a~Drop~{#}~StringStartsQ~a[[#]]&/@Range@Length@a]

PostgreSQL, 186, 173 bytes

WITH y AS(SELECT * FROM t,LATERAL unnest(c)WITH ORDINALITY s(z,r))
SELECT y.c,EVERY(u.z IS NULL)
FROM y LEFT JOIN y u ON y.i=u.i AND y.r<>u.r AND y.z LIKE u.z||'%' GROUP BY y.c

Output:

No live demo this time. http://sqlfiddle.com supports only 9.3 and to run this demo 9.4 is required.

How it works:

1. Split string array with number and name it y
2. Get all y
3. LEFT OUTER JOIN to the same derived table based on the same i(id), but with different oridinal that start with prefix y.z LIKE u.z||'%'
4. Group result based on c (initial array) and use EVERY grouping function. If every row from second table IS NULL it means there is no prefixes.

Input if someone is interested:

CREATE TABLE t(i SERIAL,c text[]);

INSERT INTO t(c)
SELECT '{"Hello", "World"}'::text[]
UNION ALL SELECT  '{"Code", "Golf", "Is", "Cool"}'
UNION ALL SELECT  '{"1", "2", "3", "4", "5"}'
UNION ALL SELECT  '{"This", "test", "case", "is", "true"}'         
UNION ALL SELECT  '{"111", "010", "000", "1101", "1010", "1000", "0111", "0010", "1011","0110", "11001", "00110", "10011", "11000", "00111", "10010"}'
UNION ALL SELECT  '{"4", "42"}'
UNION ALL SELECT  '{"1", "2", "3", "34"}'                   
UNION ALL SELECT  '{"This", "test", "case", "is", "false", "t"}'
UNION ALL SELECT  '{"He", "said", "Hello"}'
UNION ALL SELECT  '{"0", "00", "00001"}'
UNION ALL SELECT  '{"Duplicate", "Duplicate", "Keys", "Keys"}';

EDIT:

SQL Server 2016+ implementation:

WITH y AS (SELECT *,z=value,r=ROW_NUMBER()OVER(ORDER BY 1/0) FROM #t CROSS APPLY STRING_SPLIT(c,','))
SELECT y.c, IIF(COUNT(u.z)>0,'F','T')
FROM y LEFT JOIN y u ON y.i=u.i AND y.r<>u.r AND y.z LIKE u.z+'%' 
GROUP BY y.c;

LiveDemo

Note: It is comma separated list, not real array. But the main idea is the same as in PostgreSQL.

EDIT 2:

Actually WITH ORDINALITY could be replaced:

WITH y AS(SELECT *,ROW_NUMBER()OVER()r FROM t,LATERAL unnest(c)z)
SELECT y.c,EVERY(u.z IS NULL)
FROM y LEFT JOIN y u ON y.i=u.i AND y.r<>u.r AND y.z LIKE u.z||'%' GROUP BY y.c

SqlFiddleDemo

Mathematica, 41 bytes

!Or@@StringStartsQ@@@Reverse@Sort@#~Subsets~{2}&

Pyth - 13 12 bytes

!ssmm}d._k-Q

Try it online here.

Ruby, 48 bytes

Uses arguments as input and stdout as output.

p !$*.map{a,*b=$*.rotate!
a.start_with? *b}.any?

Scala, 71 bytes

(s:Seq[String])=>(for{x<-s;y<-s}yield x!=y&&x.startsWith(y)).forall(!_)

Racket 130 bytes

(define g #t)(for((n(length l)))(for((i(length l))#:unless(= i n))(when(string-prefix?(list-ref l i)(list-ref l n))(set! g #f))))g

Ungolfed:

(define(f l)
  (define g #t)
  (for ((n (length l)))
    (for ((i (length l)) #:unless (= i n))
      (when (string-prefix? (list-ref l i) (list-ref l n))
        (set! g #f))))g)

Testing:

(f [list "Hello" "World"])             
(f [list "Code" "Golf" "Is" "Cool"])
(f [list "1" "2" "3" "4" "5"])
(f [list "This" "test" "case" "is" "true"])          
(f [list "111" "010" "000" "1101" "1010" "1000" "0111" "0010" "1011" 
         "0110" "11001" "00110" "10011" "11000" "00111" "10010"])

(f [list "4" "42"])                             
(f [list "1" "2" "3" "34"])                   
(f [list "This" "test" "case" "is" "false" "t"])
(f [list "He" "said" "Hello"])
(f [list "0" "00" "00001"])
(f [list "Duplicate" "Duplicate" "Keys" "Keys"])

Output:

#t
#t
#t
#t
#t
#f
#f
#f
#f
#f
#f