Suppose that $\mathcal{A}, \mathcal{B}$ are strictly unital $A_\infty$ categories, and $\mathcal{F}, \mathcal{G}: \mathcal{A} \rightarrow \mathcal{B}$ are (strictly unital) functors.
On one hand, we can say that $\mathcal{F}$ and $\mathcal{G}$ are quasi-equivalent if the honest functors $H(\mathcal{F}), H(\mathcal{G}): H(\mathcal{A}) \rightarrow H(\mathcal{B})$ agree. (Perhaps, morally, I want to write 'up to natural isomorphism' here.)
On the other, I can consider $\mathcal{F}$ and $\mathcal{G}$ as objects of the functor category $\sf{fun}(\mathcal{A}, \mathcal{B})$. Here, I say that $\mathcal{F}$ and $\mathcal{G}$ are homotopy equivalent if they are isomorphic as elements of $H(\sf{fun}(\mathcal{A}, \mathcal{B}))$; to spell this out, this is if there are natural transformations $S: \mathcal{F} \Rightarrow \mathcal{G}$ and $T: \mathcal{G} \Rightarrow \mathcal{F}$, a pre-natural transformation $P: \mathcal{F} \Rightarrow \mathcal{F}$ with $$\mu^1_{\sf{fun}(\mathcal{A}, \mathcal{B})}(P) = T \circ S - \mathrm{Id}_F,$$ and similar for $S \circ T$, via a pre-natural transformation $Q$.
Is it true that if $\mathcal{F}$ and $\mathcal{G}$ are quasi-equivalent then they are homotopy equivalent?
If the answer is 'not in general', are there conditions upon $\mathcal{B}$ which make it so? I think, if I understand the literature, that if $\mathcal{B}$ is the (dg) category of chain complexes so that the functor category is $\sf{Mod}_{\mathcal{A}^{\mathrm{opp}}}$ then the answer is 'yes' - perhaps provided that the ground ring is a field of characteristic two?
(This is, clearly, a somewhat naiive question to ask; I ask it because of the general phenomenon in $A_{\infty}$ categories that often quasi-somethings are also homotopy-somethings, usually via some sort of perturbation theory argument - I can't see anything to the effect of the above in, e.g. Seidel's book, but I also lose my marbles a little moving between the relevant categories and so could quite easily have overlooked something.)
