I think that the answer for the first question is "yes". I disagree with the Joel's Joey's argument because he uses a wrong definition of $Comm_\infty$ morphism. Let me recall the right definition first. A $Comm_\infty$ algebra $C$ is a coderivation $Q$ such that $Q^2=0$ on cofree LIE coalgebra cogenerated by $C[1]$. So, it should be a free Lie coalgebra, not COMMUTATIVE one. Now a $Comm_\infty$ morphism is just a map of from the two cofree Lie coalgebras which commutes with $Q$. It is a quis if it is, moreover, a quis of complexes.
Well, now suppose that $C$ and $D$ are two cdga placed in any degrees. If they are $A_\infty$ quis, it means that there is a map of cofree dg coalgebra cogenerated by $C[1]$ to the same guy cogenerated by $D[1]$. This is a map of coalgebras, then it defines a map between their primitive elements.
(Recall $x$ is primitive iff $\Delta x=1\otimes x+x\otimes 1$). Apriori this gives a map of free Lie coalgebras, but NOT dg Lie coalgebras. Moreover if $C$ and $D$ would be non-commutative, the primitive elements would be not closed under the differential. But if $C$ and $D$ are commutative, the maps of the Lie coalgebras of primitive elements is a map of dg Lie coalgebras.
It is only remains to note that this map of dg Lie coalgebras is quis; then it is by definition a $Comm_\infty$ quis map. But this easily follows from the PBW theorem. The tensor coalgebra as a vector space is the symmetyric (co)algebra of the free Lie coalgebra. If the map of Lie coalgebras would be not a quis, the corresponding map of their symmetric powers also would be not a quis. This would contradict that the initial map was an $A_\infty$ quis.
We are done.
