Using your strategy, worst case scenario:
1: six and six.
2: take the heavier side, 3 and 3. Comes out even.
3: take the other six, 3 and 3. One side is lighter.
4: NOW you're down to 3. Weigh 2 of them and find the light BB.
Since the rogue BB can be either heavier or lighter, and you don't know which, you can't eliminate them that way.
Edit: I went and re-read the question, and there is some room for ambiguity in there. To re-state: One BB weighs different than the rest. You don't know if it's heavier or lighter, just that it's different.