Is the computation of a satisfying variable assignment for a Boolean formula $FP^{NP}$-hard?
$\let\mr\mathrm$Let me denote the problem as $S$. Gottlob and Fermüller state that if $S$ is solvable in $\mr{FP}^{\mr{SAT}[\log n]}$, then $\mr{P}=\mr{NP}$. However, the argument actually shows more generally that if $S$ is in $\mr{FP}^{\mr{SAT}[q(n)]}$, then SAT is solvable in time $\mr{poly}(n2^{q(n)})$. In particular, assuming the exponential time hypothesis, $S$ cannot be solved with a sublinear number of $\mr{NP}$ queries.

On the other hand, $S$ shouldn’t be hard for $\mr{FP}^\mr{SAT}$—or even $\mr{FP}^{\mr{SAT}[2]}$—under many-one search-problem reductions. For definiteness, let me assume (following Gottlob and Fermüller) that $S$ is formulated as a total search problem, required to answer “unsatisfiable” for unsatisfiable formulas. Let $L$ be a language whose $\{0,1\}$-valued characteristic function $f_L$ is reducible to $S$, that is, there are poly-time functions $g,h$ such that
$$S(g(x),y)\implies f_L(x)=h(x,y).$$
Then
\begin{align}
x\in L\iff{}&amp;\bigl(g(x)\in\mr{UNSAT}\land h(x,\text{‘unsat’})=1\bigr)\\
&amp;\lor\exists y\,\bigl(y\text{ satisfies }g(x)\land h(x,y)=1\bigr),
\end{align}
and we have a similar expression for $\overline L$ with $0$ in place of $1$, hence
$$L\in\mr{DP}\cap\mr{coDP}.$$
If this were true for all $\mr{DP}$ languages $L$, then the Boolean hierarchy would collapse to $\mr{DP}$, hence the polynomial hierarchy would collapse to something similar one level up. So, assuming no such collapse takes place, there are $\mr{FP}^{\mr{SAT}[2]}$ functions (in fact, characteristic functions of $\mr{DP}$ predicates) not reducible to $S$.

Note that if we instead formulate $S$ so that it is undefined on unsatisfiable inputs, it only becomes easier; it is a (non-total) $\mr{FNP}$ problem. Then it is not hard to see that $f_L$ is reducible to $S$ only if $L\in\mr{NP}\cap\mr{coNP}$. In particular, $S$ is not hard for $\mr{FP}^{\mr{SAT}[1]}$ (or characteristic functions of $\mr{NP}$ languages) unless $\mr{NP}=\mr{coNP}$.