Is the polynomial de Rham functor a Quillen equivalence?

This is more like a comment, but the usual space allowed for comments is too small!

It seems (after Toën's work) that we cannot describe rational homotopy types in terms of algebra, but rather in terms of (derived) algebraic geometry; the fact that commutative dg algebras (i.e. affine derived schemes) are sufficient for nice simply connected spaces is really due to the lack of monodromy. This is one of the reasons why Toën developped his theory of affine homotopy types (partly following ideas which were already sketched in Grothendieck's Pursuing stacks): for any ring $k$ and any space $X$, there is the pro-unipotent completion of the $\infty$-groupoid corresponding to $X$, denoted by $(X\otimes k)^{uni}$, which is the higher analog of the pro-unipotent completion of a group. If $k$ is a field of characteristic zero, and if $X$ is $1$-connected, nilpotent and of finite type, then the stack $(X\otimes k)^{uni}$ is the spectrum (in the sense of derived algebraic geometry) of the commutative dg algebra of de Rham cohomology of $X$, so that this theory includes classical rational homotopy theory. The essential information given by the stack $(X\otimes k)^{uni}$ is essentially the $\infty$-category of $k$-linear local systems over $X$.

All this is explained in the paper

B. Toën, Champs affines, Selecta Math. (N.S.) 12 (2006), no. 1, 39-135.

(see in particular Cor. 2.4.11, Cor. 2.5.3 and Cor. 2.5.4, to see that this extends nicely classical rational homotopy theory to non-simply connected homotopy types).

The answer to question 1 is no. For one thing, the functor to the category of commutative DGAs is not essentially surjective. For any space $X$, $H^0(X)$ is a product of copies of $\mathbb Q$, whereas it is relatively easy to rig up CDGAs that don't have this property.

More seriously, nonnegatively (cohomologically!) graded CDGAs don't form a model category using the structure you've listed. Try to construct a cofibrant replacement for $\mathbb Q[x]/x^2$ concentrated in degree zero, and you'll find that you want to adjoin classes in degree $-1$; being cofibrant means, by consideration of maps to DGAs concentrated in degree zero, that the algebra in degree zero has a strong lifting property that can't be altered.

One thing to note about question 2 is that the functor $\pi_0$ is still well-defined on the rational homotopy category, because rational equivalences are always isomorphisms on $\pi_0$. Any algebraic model that you construct, then, has to account for this. (This is not even beginning to worry about $\pi_1$, which always messes things up.)