I don't know the full answer but it's not true that if $A$ and $B$ are quasi-isomorphic then so are $Der(A)$ and $Der(B)$. The following is over $\mathbb Q$. The simplest counterexample is the minimal model of $S^2$. $S^2$ is formal so its minimal model is quasi-isomorphic to the one for its cohomology. Note that $H^*(S^2)$ has no odd derivations being evenly graded. Yet, its minimal model is $M= (\Lambda (x,y), d)$ with $ deg (x) =2, deg (y)= 3$ and $dx=0, dy=x^2$. You can directly see that $H_{odd}(Der(M))$ is not zero. Specifically, there is a closed non cohomologous to zero derivation $D$$\theta$ of degree $-3$ with $D(x)=0, D(y)=1$$\theta(x)=0, \theta(y)=1$. In In general there is a natural map $H(Der(M))\to Der(H(M))$ which is onto for formal spaces but it need not be injective. This This map looks like it should be the edge homomorphism in some spectral sequence as it commutes two functors (which is common for some natural spectral sequences like the Eilenberg-Moore spectral sequence for diff Tor) but I don't know what that spectral sequence should be.
It It is however true that for positively elliptic spaces such as $S^2$ (they are all formal) one has that $H_i(Der(M))\to Der_i(H(M))$ is an isomorphism for negative even $i$.
Here Here a space is called (rationally) elliptic if it has finite dimensional total rational cohomology and homotopy. Basic examples are homogeneous spaces and fiber bundles built out of them. A space is called positively elliptic if it's elliptic and has positive Euler characteristic (e.g. $S^2$). This is equivalent to saying that its homotopy Euler characteristic is zero, i.e. the total rank of all even homotopy groups is the same as the total rank of odd homotopy groups. See the book by Félix, Halperin and Thomas on the basics of elliptic spaces.[Félix, Halperin and Thomas on the basics of elliptic spaces.][1] [1]: http://www.ams.org/mathscinet/search/publdoc.html?r=1&pg1=CNO&s1=1802847&loc=fromrevtext