Quantum complexity of TQBF with an untrusted oracle
This question is more complicated to state than its predecessor Quantum complexity of TBQF, but in hindsight it is also way easier to answer.
Theorem: The complexity is $\tilde{O}\left(2^{n/4}\right)$, using a single call to Grover's algorithm to check if there is a path falsifying the oracle.  Moreover, this is optimal in the black box case.
Consider a TBQF query $F = \exists x_0 \forall y_0 \exists x_1 \forall y_1 \cdots f(x, y)$.  Assume w.l.o.g. that the oracle says $F$ is true.  The oracle is telling the truth about $F$ if, when we use the oracle to choose how to play each $x_k$ move, there is no choice of $y$'s that falsifies the oracle.  But this is just a single call to Grover to search through the space of such $y$'s.  There are $2^{n/2}$ $y$'s, so the complexity is $\tilde{O}\left(\sqrt{2^{n/2}}\right) = \tilde{O}\left(2^{n/4}\right)$.
For the black box lower bound, given a black box boolean function $f(y)$ defined on $n/2$ bits $y$, we can construct a degenerate TBQF formula $F = \exists x_0 \forall y_0 \exists x_1 \forall y_1 \cdots f(y)$ which simply ignores $x$.
(Thanks to Paul Christiano for discussion.)