I guess one option would be to just declare that all NaNs are equal--I'm pretty sure that's how bottoms work in Haskell, and it seems that NaN is essentially a floating-point version of bottom.
This is somewhat simpler--as long as either argument to (==) is NaN the answer is False. However, you still have to figure out that at least one bit pattern corresponds to NaN. If you want them to be equal, you would have to figure out that both are NaN. This is certainly a little more difficult, but I think it isn't much worse than the current scenario.