Suppose we have Grothendieck abelian categories $\mathcal{A}, \mathcal{B}$. Suppose also we have given an exact functor of triangulated categories $$ F \colon D(\mathcal{A}) \to D(\mathcal{B}) $$ where $D(\mathcal{A})$ and $D\mathcal({B})$ denote appropriate derived categories of complexes (possibly bounded below or both ways). Suppose that for any injective $I \in \mathcal{A}$ we have a functorial maps which is a quasi-isomorphism $$ F(I) \cong h^0F(I). $$ Is it then true that $F$ is the right derived functor of is zeroth cohomology? I.e. $F \cong R(h^0F)$. If not, are there known counterexamples? Or which approporiate additional assumptions are needed so that such a statement holds?

I believe that the above setup yields that the $i$th cohomology of $F$ is canonically isomorphic to the $i$th right derived functor of $h^0F$. However, I don't see how to extend the given quasi-isomorphism $F(I) \cong h^0F(I)$ for injectives in $\mathcal{A}$ to a natural transformation of functors $F \circ Q \to h^0F$ where $Q$ denotes the natural functor from the homotopy category $K(\mathcal{A})$ to $D(\mathcal{A})$. Once one has this, the universal porperty of the right derived functor should give the result.

Some Background: In the situation I am interested in the functor $F$ arises as a composition of a left and right derived functor in a much bigger ambient category, whose restriction to $D(\mathcal{A})$ happens to be left exact and satisfy the condition $F(I) \cong h^0F(I)$ for injective objects of $\mathcal{A}$. I could imagine that this is not an uncommon situation...