Let $\mathscr A$ be a Grothendieck abelian category. Then every object in $\operatorname{Ch}(\mathscr A)$ is quasi-isomorphic to a $K$-injective object [Stacks, Tag 079P]. In particular, for any left exact functor $F \colon \mathscr A \to \mathscr B$ of abelian categories, the derived functor $RF \colon D(\mathscr A) \to D(\mathscr B)$ is everywhere defined.
However, we rarely ever compute $RF(C^\bullet)$ using $K$-injective resolutions. Instead you use spectral sequences (or basically any tool you can get your hands on). While I don't expect methods in complete generality for computing $RF(C^\bullet)$, there is a relatively mild case that I also couldn't figure out:
Question. If $C^\bullet \in \operatorname{Ch}(\mathscr A)$ is a complex such that $RF(Z^i(C^\bullet)) = RF(B^i(C^\bullet)) = 0$ for all $i$, then is it true that $RF(C^\bullet) = 0$?
I don't know a name for the condition that $RF(A) = 0$ for $A \in \mathscr A$, but let me say that $A$ is totally $F$-acyclic in this case. For instance, if $\mathscr A = \operatorname{Sh}(X)$ for $X = [0,1]$ and $U = [0,1)$, then $\mathbf Z_U$ is totally $F$-acyclic for $F = \Gamma(X,-)$ since $R\Gamma(X,\mathbf Z) \to R\Gamma(X,\mathbf Z_{X\setminus U})$ is an isomorphism (that is, $R\Gamma_c([0,1),\mathbf Z) = 0$).
Note that the hypotheses imply that $C^i$ and $H^i(C^\bullet)$ are totally $F$-acyclic as well, via the short exact sequences \begin{align*} 0 \to Z^i \to\ \!& C^i \to B^{i+1} \to 0, \\ 0 \to B^i \to\ \!& Z^i \to H^i\ \to 0. \end{align*} The case I am most interested in is $\mathscr A = \operatorname{Sh}(X)$ on some reasonable site (e.g. a locally compact Hausdorff space) and $F = \Gamma(X,-)$. As suggested in Jeremy Rickard's answer, we should at least assume that $\mathscr B$ has exact (countable) products and $F$ preserves (countable) products (these hypotheses also feature in [Tag 08U1]).
Example. If $C^\bullet$ is bounded below, then [Stacks, Tag 015W] gives spectral sequences \begin{align*} E_1^{p,q} &= R^qF(C^p) & \Rightarrow R^{p+q}F(C^\bullet)\\ E_2^{p,q} &= R^pF(H^p(C^\bullet))\hspace{-1.8em} & \Rightarrow R^{p+q}F(C^\bullet) \end{align*} using the filtrations $F^pC^\bullet = \sigma_{\geq p}C^\bullet$ and $F^pC^\bullet = \tau_{\leq -p}C^\bullet$ respectively (in the latter case there is a re-indexing – see [Stacks, Tag 0FLJ] for details when $F = \Gamma(X,-)$ for sheaves on a topological space $X$). Both spectral sequences vanish on their first page, so the result follows from convergence of these spectral sequences. In particular, we may assume $C^\bullet$ is bounded above by the short exact sequence $0 \to \sigma_{\geq 1}C^\bullet \to C^\bullet \to \sigma_{\leq 0} C^\bullet \to 0$ and the result for $\sigma_{\geq 1} C^\bullet$.
The general version of this spectral sequence argument [Stacks, Tag 0BK5 or 0BKK] assumes that you already know that $RF^n(F^p C^\bullet) = 0$ for $p \gg 0$ and $RF^n(F^p C^\bullet) \stackrel\sim\to RF^n(C^\bullet)$ for $p \ll 0$, which you usually only verify by using filtrations that are bounded on one side. In my situation, you cannot verify these hypotheses if you don't already know vanishing of $RF^n(\sigma_{\leq p-1} C^\bullet)$ for $p \ll 0$ (for the first filtration) or of $RF^n(\tau_{\leq -p} C^\bullet)$ for $p \gg 0$ (for the second one). This is another instance of the question we started with!
However, the result does follow if you also know that $C^\bullet \to \underset{\longleftarrow}{\operatorname{holim}} \tau_{\geq -n} C^\bullet$ is an equivalence, since the hypothesis on $C^\bullet$ is preserved on $\tau_{\geq -n} C^\bullet$ and $RF$ preserves homotopy limits since it preserves countable products [Stacks, Tag 08U1]. But in general, I don't see a reason why this should hold if we merely assume total $F$-acyclicity.
Remark. In the bounded below case, it is enough to assume that all $C^i$ are totally $F$-acyclic, by the first of the spectral sequences above. This is not true in the unbounded case: if $Q = [0,1]^{\mathbf N}$ is the Hilbert cube with opens \begin{align*} U_i = (0,1)^{\mathbf N_{<i}} \times [0,1) \times [0,1]^{\mathbf N_{>i}},\\ V_i = (0,1)^{\mathbf N_{<i}} \times (0,1] \times [0,1]^{\mathbf N_{> i}}, \end{align*} then $Q = U_0 \cup V_0$ and $U_i \cap V_i = U_{i+1} \cup V_{i+1}$ for all $i$, leading to a resolution $C^{-i} = \mathbf Z_{U_i} \oplus \mathbf Z_{V_i}$ of $\mathbf Z$. All $R\Gamma(Q,\mathbf Z_{U_i}) = R\Gamma_c(U_i,\mathbf Z)$ vanish, and likewise for $V_i$, but $R\Gamma(Q,C^\bullet) = R\Gamma(Q,\mathbf Z) = \mathbf Z[0]$. See also this answer where this resolution plays a role.
I don't know a similar example if we only assume that all $H^i(C^\bullet)$ are totally $F$-acyclic, so that could be another (strictly harder) question.
References.
[Spalt] N. Spaltenstein, Resolutions of unbounded complexes. Compos. Math. 65.2, p. 121-154 (1988).
[Stacks] The Stacks project.
