Consequences of $BQP \subseteq P/poly$?
I'd say we have no good reason to think BQP is in P/poly.  We do have reasons to think that BQP is not in P/poly, but they're more-or-less identical to our reasons to think that BQP&ne;BPP.  E.g., if BQP&sub;P/poly then Factoring is in P/poly, which is enough to break lots of cryptography according to standard security definitions.

Also, as you correctly point out, there's no quantum analogue of Adleman's trick---indeed, there's no way to "pull the quantumness out of a quantum algorithm," analogous to how one can pull the randomness out of a randomized algorithm.  So I don't think anyone has a guess for what the P/poly advice for simulating a quantum computer should even consist of (any more than they have a guess, say, in the case of NP vs. P/poly).

A final note: my work with Alex Arkhipov (and the independent work of Bremner-Jozsa-Shepherd), can easily be adapted to show that if QUANTUM-SAMPLING is in P/poly (OK, in "BPP-SAMPLING/poly"), then P#P&sub;BPPNP/poly, and hence the polynomial hierarchy collapses---in this case, I think, to the fourth level.  At present, though, we don't know how to adapt this sort of result from sampling problems to decision problems.