If BQP is contained in any level of the Polynomial Hierarchy, does it then follow that $NP \subseteq BQP$ implies $PH \subseteq BQP$?

Question

I think this is implied in this paper by Aaronson (http://www.scottaaronson.com/papers/bqpph.pdf) but I am not sure.

Begin with $NP \subseteq BQP$ (*)

$\Sigma_{2}^{P} = NP^{NP} \subseteq BQP^{BQP} = BQP$

$\Sigma_{i+1}^{P} = NP^{\Sigma_{i}^{P}} \subseteq BQP^{BQP} = BQP$

With this inductive step we have $PH \subseteq BQP$

The only part of this proof I doubt is the step $NP^{NP} \subseteq BQP^{BQP}$ because I am unsure of some oracle rules.

My assumption is that if $A \subseteq B$ then:

$A^{C}\subseteq B^{C}$

and

$C^{A}\subseteq C^{B}$

with no assumptions about $C$, are these correct?

I feel like this would be a well-known result if it were true without any additional assumptions, so my belief is that I must need to make an additional assumption at some point that I have missed.