I can say that if A and B are commutative DGAs over Q which are connective (their homology groups are zero in negative degrees, using homological grading), then A and B are equivalent. The only proof I can come up with is a Postnikov argument based on the Hochschild-Kostant-Rosenberg theorem.

Namely, let P_{n}A and P_{n}B be Postnikov stages, which have the same homology as A and B respectively in degrees less than or equal to n and zero above; you can construct these by adding new polynomial generators to A and B with differentials that erase higher homology classes. Assume you've already constructed an equivalence from P_{n}A to P_{n}B.

Then P_{n+1}A and P_{n+1}B are classified by their next k-invariants. In the associative case this is an element in the Hochschild cohomology of P_{n}A with coefficients in a shift of the module M = H_{n+1}A = H_{n+1}B, and similarly for B. In the commutative case this is an element in the Andre-Quillen cohomology instead. These are classified by maps from the Hochschild homology (strictly speaking, its augmentation ideal) and Andre-Quillen homology (a shift of the derived Kahler differentials) to this shift of M.

Since we're in characteristic zero, the HKR theorem tells us that the Hochschild homology is the (derived) free graded-commutative algebra over P_{n}A on the Andre-Quillen homology. In particular, there is a retraction down. This implies that the collection of possible commutative k-invariants constructing an extension of P_{n}A by M are a *subset* of the possible associative k-invariants.

As a result, if P_{n+1}A and P_{n+1}B have a fixed equivalence as associative algebras, that implies that their associative k-invariants are equal. Hence that their commutative k-invariants are equal, so this allows us to lift to an equivalence of commutative algebras.

This is a sneaky problem. The fact that I needed to assume connectivity for this argument is pretty annoying, and I am not sure whether it is material or simply a failing on my part to get something general. Perhaps someone else can do better.

ADDED LATER: I believe that the same basic argument works in the coconnective case under added hypotheses (e.g. the algebra being simply connected, having only Q in degree 0 and 0 in degree 1). Unfortunately you cannot construct Postnikov stages quite as easily for general cochain algebras.

For example, let A be the commutative DGA Q[x,y,s,t] ⊗ Λ[w], a free algebra on generators x,y,s,t in degree zero and w in (cohomological) degree 1, satisfying dx = dy = dw = 0, ds = xw, dt = wy. Then there is a Massey product `<x,w,y>`

= sy - xt in cohomology degree 0 which is nonzero, and any map from A to a DGA which has zero cohomology in positive degrees must necessarily send w to zero, and hence also this Massey product.

I have this vague feeling that in this simply-connected case (and also in the connective case) there might be a much shorter argument involving minimal models.