Resolution of a module as an $A_\infty$ module over resolution of an algebra The following question looks like a basic question on resolutions over commutative rings, unfortunately I was not able to find a reference.
Let $A$ be a regular commutative noetherian ring  (and satisfy all other nice properties that could be needed for construction) , $I \subset A$ is an ideal. Suppose $D$ is a differential graded algebra resolution of $A/I$ over $A$ i.e. $D$ is a free resolution of $A/I$ that admits a structure of dg-algebra over $A$. Basic example of such resolutions is a Koszul complex for ideals generated by a regular sequence. Now let $M$ be a finitely generated module over $A/I$, and $F$ is a free resolution of $M$ over $A$. Is it true that that $F$ is an $A_\infty$-module over $D$?
It is well known that there is a resolution with a dg-module structure over $D$, my question is that any resolution $F$ admits an $A_\infty$-module structure over $D$. 
 A: I'll try to endow $F$ with a strict $D$-algebra structure. The endomorphism DG-$A$-algebra $\operatorname{End}_A(F)$ satisfies
$$H_n\operatorname{End}_A(P)=\operatorname{Ext}^{-n}_A(M,M).$$
In particular the homology vanishes in positive dimensions. A $D$-algebra structure on $F$ is an map of DG-$A$-algebras $D\rightarrow \operatorname{End}_A(F)$. Since $D$ is concentrated in non-negative degrees, this is the same as a map to the truncation $D\rightarrow t_{\geq 0}\operatorname{End}_A(F)$, whose homology is $\operatorname{Hom}_A(M,M)$ concentrated in degree $0$.Therefore, the natural projection $t_{\geq 0}\operatorname{End}_A(F)\twoheadrightarrow\operatorname{Hom}_A(M,M)$ is a quasi-isomorphism, indeed a trivial fibration in the projective model structure on the category of DG-$A$-algebras.
Since $M$ is actually an $A/I$-module, the 'enriched identity in $M$', which is given by an $A$-module morphism $A\rightarrow\operatorname{Hom}_A(M,M)$, factors through the natural projection $A\twoheadrightarrow A/I$.
Now, consider the following commutative square in the category of DG-$A$-algebras, where $A$ is the initial object, 
$$\begin{array}{ccccc}
A&-&-&\rightarrow&t_{\geq 0}\operatorname{End}_A(F)\\
\downarrow&&&&\downarrow\\
D&\stackrel{\sim}\longrightarrow&A/I&\rightarrow&\operatorname{Hom}_A(M,M)
\end{array}$$
The upper arrow should be continuous, but I didn't manage to get it. The right vertical arrow is a trivial fibration, as I've already noticed. The map $A\rightarrow D$ should be a cofibration (this is how I understand that $D$ is a resolution of $A/I$) and $D\rightarrow A/I$ is the resolution map (a trivial fibration, or just an equivalence, it doesn't really matter).
Then, the model category axioms give you a map $D\rightarrow t_{\geq 0}\operatorname{End}_A(F)$ such that, when fitted in the previous square, the two triangles commute.

After your clarification, I can complete my argument to give a positive answer to your question. Before $D$ was a DGA resolution of $A/I$. Now I simply take 
a DGA $D'$, levelwise projective, concentrated in non-negative degrees, and with $H_*(D')=A/I$ concentrated in degree $0$. The natural projection $D'\twoheadrightarrow A/I$ is therefore a trivial fibration. Hence, the following commutative diagram has a lift by the same argument as before
$$\begin{array}{ccc}
A&\rightarrow &D'\\
\downarrow&&\downarrow\\
D&\rightarrow &A/I
\end{array}$$
The lift $D\rightarrow D'$ is a quasi-isomorphism by the 2-out-of-3 property, since $D\rightarrow A/I$ and $D'\rightarrow A/I$ are quasi-isos. Hence, it has an up-to-homotopy inverse $D'\rightarrow D$ which is an A-infinity morphism. It is exactly here where the fact that the underlying complex of $D'$ is a projective resolution is used.
The composite $D'\rightarrow D\rightarrow t_{\geq 0}\operatorname{End}_A(F)\subset \operatorname{End}_A(F)$ is an A-infinity morphism which endows $F$ with the structure of an A-infinity $D'$-module such that the induced $A/I$-module structure on $M$ (obtained by taking $H_0$) is the given one.
