Python 2: 58 chars
n=input() s="" while n:s="0+-"[n%3]+s;n=-~n/3 print s or 0
Generates the balanced ternary digit-by-digit from the end. The last digit is given by the residue
+1. We then remove the last digit and divide by 3 using Python's floor-divide
n=(n+1)/3. Then, we proceed recursively with the new last digit until the number is 0.
A special case is needed for the input
0 to give
0 rather than the empty string.
The specs don't allow this, but if one could write a function instead of a program and output the empty string for 0, a 40 char solution would be possible.
g=lambda n:n and g(-~n/3)+"0+-"[n%3]or""