I'm not a crypto master, but as I understand it, while it's possible to find md5 collision, it's many times more difficult in hmac, because it's:

    hash((secret+pad1) | hash((secret+pad2) | message))

so you would have to find a collision of one key that matches collision of another key, so you're back to relying on basic birthday attack rather than any specific hash weakness.

If you're looking for an actual proof, it's at http://cseweb.ucsd.edu/~mihir/papers/hmac-new.html