Let $M$ and $N$ be homotopical categories (both of them complete) and $I$ a small category. Assume that we have right deformations for the limit functors $M^I \to M$ and $N^I \to N$, so we have the (non-total) right derivatives \begin{align} \text{holim} : M^I &\to M \\ \text{holim} : N^I &\to N.\end{align} Assume now that we have a right-adjoint functor $F: M \to N$ (so it commutes with the limit) that is additionally homotopical, i.e. it preserves all weak equivalences. Question: Is there a natural weak equivalence between $\text{holim} \circ F$ and $F\circ \text{holim}$? In other words: Will $F$ (applied to diagrams) also preserve homotopy limits? If this is not true, what might be additional requirements on $F$ to make it true?
Thank you for any hints.
Addendum: I have a little more concrete problem in mind, in which $M$ is given by non-negative cochain complexes and $N$ by non-negative chain complexes. I think I could perform a proof by direct computation if I can find concrete descriptions of the cotensors in these simplicial model categories. I only found them for (co)simplicial Abelian groups and I am a little unsure how to transfer them to complexes. Has anyone seen that written down explicitly?
The tensors and cotensors in chain complexes come via the Moore complex realization of simplicial sets. For instance, the Moore complex of the interval over a commutative ring $R$ is $...\to 0\to R\to R\oplus R\to 0\to...$ where the nontrivial map is $(1,-1)$. The cotensor with the interval, that is, the path object, is the complex of maps out of this complex. In general, you can write down a formula for any cotensor as a limit of cotensors by simplices, although that might not be explicit enough for your needs.
For your abstract problem: let the deformations be $R_M, R_N$. Then $$F(\mathrm{holim}D)=F( \mathrm{lim}(R_M(D)))\cong \mathrm{lim}F\circ R_M(D)...\mathrm{lim}R_N F\circ D=\mathrm{holim} F\circ D$$ So we see the problem is concentrated in that ellipsis. We can fill it in using the weak equivalences $r_D:D\to R_MD,r_{F\circ D}:F\circ D\to R_N(F\circ D)$, but only as a zigzag.