Just another way to do it without loops (kind of cheating):

my $c=0;
s/ \(      (?{++$c})
 | \)      (?{--$c})
 | ([^()]+)
 /$1 if ($c==0)/gex;
[download]

If you match an open paren, increment a counter;
if you match a close paren, decrement it;
if you match a string of non-parens, capture it;
substitute whatever was captured back in if the counter is zero.

There is an "if", but it can be replaced by an "x" operator, if you're a stickler. Or you could rewrite the replacement as $c==0 and $1.

Another little variation that doesn't rely on the (?{}) construct:

s{([()]*)([^()]+)}
 {$c += $1=~y/(// - $1=~y/)//;
  $2 x !$c
 }ge;
[download]

Match any parens followed by any non-parens;
Add the count of opens and subtract the count of closes from your counter;
Substitute the non-paren portion back in if the counter is zero.