This question loosely elaborates on an earlier question. It is pretty silly, but I'd like to hear some authoritative answers.

Recall that if $f:S^{\bullet}\to T^{\bullet}$ is a quasi-isomorphism of sheaves over $X$, which is, say, a manifold, then for every open set $U\subset X$, we have an induced isomorphism $R\Gamma(U,S^{\bullet})\to R\Gamma(U,T^{\bullet})$, so $H^i(U,S^{\bullet})\cong H^i(U,T^{\bullet})$ and in particular $H^i(X,S^{\bullet})\cong H^i(X,T^{\bullet})$.

To what extent is the converse true? At the coarsest level, when does a canonical isomorphism $R\Gamma(X,S^{\bullet})\to R\Gamma(X,T^{\bullet})$ reflect an underlying derived equivalence?

For a counterexample to the coarsest case, I believe the following serves: Consider a space $X$. Consider the constant sheaf on $k_X$. Let $f:X\to x_0$ be the retraction to a point $x_0\in X$. By standard theorems, we know that $H^i(X,Rf_*k_X)\cong H^i(X,k_X)$, but evaluating $Rf_*k_X(U)$ on any open subset $U$ missing $x_0$ assigns zero, as the fiber is empty. So in general these sheaves are not derived equivalent. What if $X$ deformation retracts to $x_0$? Is $k_X$ and $Rf_*k_X$ derived equivalent then? What if the homotopy doesn't have some Vietoris-Begle type behavior? See Kashiwara Schapira 2.7.8.