**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 $\mathscr{C}$.