When is a functor a right derived functor?

Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories with enough injective objects. Let me use the notation $D^+(\mathcal{A})$ and $D^+(\mathcal{B})$ to denote the stable $\infty$-categories whose homotopy categories are the (cohomologically bounded below) derived categories of $\mathcal{A}$ and $\mathcal{B}$, respectively (you can also consider unbounded derived categories, but the situation is a bit more subtle).

Let $\mathcal{C} \subseteq \mathrm{Fun}( D^{+}( \mathcal{A} ), D^{+}( \mathcal{B}) )$ be the full subcategory spanned by those functors which are exact, left t-exact, and carry injective objects of $\mathcal{A}$ into the heart of $D^{+}( \mathcal{B} )$. Then the construction $$F \in \mathcal{C} \mapsto h^0 F|_{ \mathcal{A} }$$ determines an equivalence from $\mathcal{C}$ to the category of left exact functors from $\mathcal{A}$ to $\mathcal{B}$. The inverse of this equivalence is "taking the right derived functor".

Consequently, one can answer your question as follows: given a functor of triangulated categories $G: hD^{+}(\mathcal{A}) \rightarrow hD^{+}(\mathcal{B})$, it arises as a right derived functor (of a left exact functor of abelian categories) if and only if

a) The functor $G$ lifts to an exact functor of stable $\infty$-categories $D^{+}(\mathcal{A}) \rightarrow D^{+}(\mathcal{B})$ (anything that you build by composing derived functors will have this property).

b) The functor $G$ is left t-exact and carries injective objects of $\mathcal{A}$ into the heart of $hD^{+}(\mathcal{B})$.

Here's a counterexample for unbounded derived categories (this doesn't answer the revised question with the "$t$-left exact" condition).

Suppose $\mathcal{A}$ is a Grothendieck category with enough projectives, where projectives and injectives coincide: for example, the module category of $k[x]/(x^2)$ for a field $k$.Let $F$ be the left derived functor of a right exact, but not exact, functor.

Added (19 July 2016):

I don't have a definitive answer for the general revised question, but here's a reason that I think that (if the answer is "yes") then it might be hard without extra conditions.

I believe that it is even still an open problem whether there could be a self-equivalence $F$ of the derived category of a module category $\text{Mod}(A)$ such that the restriction of $F$ to $\text{Mod}(A)$ is the identity, $F(X)\cong X$ for every object $X$ of the derived category, but $F$ is not isomorphic to the identity functor. If such a functor existed, then it would of course be a counterexample.

This is something that annoyed me for several weeks when I was doing my PhD thirty-odd years ago. I made some comments in Section 7 of my paper "Morita theory for derived categories", although I'm afraid the comments don't say much more than "I don't know".

The difficulty is probably just an artefact of the axioms for an abstract triangulated category not being rigid enough, and it's likely that the problem is easier for functors that arise in nature.

Also, I don't rule out the possibility of simpler, more natural counterexamples for the question asked in the OP, although I haven't been able to think of any.