Applying F gives a termwise G-acyclic complex

Each $I^q$ is injective, so the hypothesis gives $R^{p}G(F(I^q))=0$ for $p>0$. Additivity makes $F(I)$ a cochain complex, zero in negative degrees. Its cohomology is precisely $R^{q}F(A)$ by definition. In degree zero left exactness identifies its kernel with $F(A)$; positive exactness would additionally require every positive $R^{q}F(A)$ to vanish.

For a witness take $F={\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,-)$, $G$ the identity of abelian groups and $A=\mathbb{Z}$, with a supplied injective resolution of $A$. Identity is exact, so all its positive derived objects vanish. The projective resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$ has rank-one free, hence projective, terms: a map from $\mathbb{Z}$ lifts across an epimorphism by lifting the image of $1$. Applying $\mathrm{Hom}(-,\mathbb{Z})$ gives $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ in degrees zero and one. Its degree-one cohomology is $\mathbb{Z}\mathrm{/}2\mathrm{\ne }0$. F4 identifies this with $H^1(F(I))$. Thus $F(I)$ is not a resolution of $F(A)=0$, even though every one of its terms is $G$-acyclic. This witness is relative to supplied data and uses no choice of an infinite family of lifts.