First I'll give a counterexample that shows that some extra conditions on $F$ are necessary, beyond just being left exact.
Let $A=k[x]/(x^2)$, and let $\mathscr{A}$ be the category $\operatorname{Mod}A$ of $A$-modules. Let $\mathscr{F}$ be the Serre subcategory of finitely generated modules, $\mathscr{A}/\mathscr{F}$ the quotient category, and $F:\mathscr{A}\to\operatorname{Mod}k$ the functor $\operatorname{Hom}_{\mathscr{A}/\mathscr{F}}(V,-)$, where $V$ is an infinite direct sum of copies of $A/Ax$.
Let $C^\bullet$ be the complex with $A/Ax$ in every degree, and zero differentials. Then $Z^i(C^\bullet)$ and $B^i(C^\bullet)$ have injective resolutions by objects of $\mathscr{F}$ for all $i$, and are therefore "totally $F$-acyclic".
But $C^\bullet$ has a $K$-injective resolution $J^\bullet$ that is the product of all the shifts $\{I^\bullet[t]\mid t\in\mathbb{Z}\}$ of a minimal injective resolution $I^\bullet$ of $A/Ax$, and $F(J^\bullet)$ has zero differentials, but is nonzero in every degree, so $RF(C^\bullet)\neq0$
Second, I'll give a proof assuming that $\mathscr{B}$ has exact countable products, and that $F$ preserves countable products as well as being left exact.
Since the product in $D(\mathscr{A})$ is given by taking the usual (termwise) product of $K$-injective resolutions, and countable products in $D(\mathscr{B})$ are the usual (termwise) products, it follows that the right derived functor $RF:D(\mathscr{A})\to D(\mathscr{B})$ also preserves countable products
Let $\mathscr{C}$ be the class of totally $F$-acyclic objects of $D(\mathscr{A})$: i.e., objects $X$ such that $RF(X)=0$. Then $\mathscr{C}$ is a triangulated subcategory of $D(\mathscr{A})$ closed under countable products, that by assumption contains all of the $Z^{i}(C^{\bullet})$, $B^{i}(C^{\bullet})$, $H^{i}(C^{\bullet})$ and $C^{i}$.
The brutal truncation $\sigma_{\geq0}C^{\bullet}$ is in $\mathscr{C}$ since it is the homotopy limit of its brutal truncations $\sigma_{\leq n}\sigma_{\geq0}C^{\bullet}$, which are all bounded complexes with terms in $\mathscr{C}$.
The brutal truncation $\sigma_{<0}C^{\bullet}$ is in $\mathscr{C}$ since it is the homotopy limit of its civilized truncations $\tau_{\geq -n}\sigma_{<0}C^{\bullet}$, which are all bounded complexes with terms in $\mathscr{C}$.
Hence, because of the triangle $$\sigma_{\geq0}C^{\bullet}\to C^{\bullet}\to\sigma_{<0}C^{\bullet}\to\sigma_{\geq0}C^{\bullet}[1],$$ $C^{\bullet}$ is in $\mathscr{C}$.