Is BQP equal to BPP with access to an Abelian hidden subgroup oracle? Is BQP equal to BPP with access to an Abelian hidden subgroup oracle?
Like many complexity-class separations, our best guess is that the answer is that BPP^{HSP} != BQP, but we can only prove this rigorously relative to oracles.  This separation was observed by Scott Aaronson in this blog post
where he observed that the welded-tree speedup of Childs, Cleve, Deotto, Farhi, Gutmann and Spielman was not contained in SZK.

On the other hand, BPP^{HSP} is contained in SZK, at least if the goal is to determine the size of the hidden subgroup.  This includes even the abelian HSP, although I'm not sure how exactly to find the generators of an arbitrary hidden subgroup in SZK.  The reason we can decide the size of the hidden subgroup is that if f:G->S has hidden subgroup H, and we choose g uniformly at random from G, then f(g) is uniformly random over a set of size |G|/|H|. In particular, f(g) has entropy log|G| - log|H|.  And entropy estimation is in SZK.