UPDATE after some thinking (and without comments or answers) I want to ask the following:
Say that I have a morphism $f:X\to Y$ of schemes and a induced morphism $f_*\colon \mathcal{S}(X)\to \mathcal{S}(Y)$. Let $A,B\in \mathcal{S}(X)$. Since $Rf_*$ is a functor from the derived category of $\mathcal{S}(X)$ to the derived category of $\mathcal{S}(Y)$ there is an induced morphism
$Rf_*(\textbf{Hom}(A,B))\to \textbf{Hom}(Rf_*A,Rf_*B)$. Here $\textbf{Hom}$ is the set of Homomorphisms in the derived category, we also have $H^0\textbf{RHom}=\textbf{Hom}$.
Now if we let $B=f^!A'$ for some $A'$ in the derived category of $\mathcal{S}(Y)$ the Grothendieck trace map induces a morphism $$\textbf{Hom}(Rf_*A,Rf_*f^!A')\to \textbf{Hom}(Rf_*A,A')$$
Is this composition an isomorphism?
I.e., is $$Rf_*(\textbf{Hom}(A,f^!A'))\to \textbf{Hom}(Rf_*A,A')$$ an isomorphism?
By Grothendieck duality $$Rf_*\textbf{RHom}(A,f^!A')\to \textbf{RHom}(Rf_*A,A')$$ is an isomorphism but if we apply $H^0$ to this we get
$$H^0Rf_*\textbf{RHom}(A,f^!A')\to \textbf{Hom}(Rf_*A,A')$$
Which seems pretty far from what I wanted unless if $f_*$ is exact.
Thank you for any input!