Your theorem needs at least one further assumption. Otherwise, we can let F be the functor on spaces sending $X$ to the commutative DGA $H^*(X;R)$, with zero differential.
I'm not sure what statement to add. The thing I initially wanted to write is that your functor should take homotopy pushouts to homotopy pullbacks. However, commutative DGAs aren't closed under homotopy pullback.
EDIT: Let's suppose you make this assumption. Take the diagram $* \leftarrow \mathbb{CP}^\infty \rightarrow *$, form the homotopy pushout $\Sigma \mathbb{CP}^\infty$, and apply your functor $F$ to the associated square pushout diagram. You get a commutative square of commutative differential graded algebras, which is in particular a commutative square of $E_\infty$-algebras. The category of $E_\infty$-algebras has homotopy pullbacks, and so you can construct by this method a zigzag
$weak equivalence $F(\Sigma \mathbb{CP}^\infty) \leftarrow C \rightarrow P
$$
of weak equivalences P$ of $E_\infty$-algebras, where $C$ is a cofibrant replacement and $P$ is the homotopy pullback.
However, in any characteristic the Steenrod operations are stable operations and an invariant of weak equivalence of $E_\infty$-algebras. If $pR = 0$, the element $x \in H^2(F(\mathbb{CP}^\infty))$ supports nonzero power operations (namely, powers) at any prime, and so the associated element
$$\sigma x \in H^3(P) \cong H^3(F(\Sigma \mathbb{CP}^\infty))$$
would also support nonzero power operations. But for commutative DGAs those operations are automatically zero.
Ideally one could make this for more general $R$ where $p$ is not invertible using a secondary operation and $B\mathbb{Z}/p$ instead, but the kids were up early and I'm too tired to figure out how right now.
