@HarryGindi That's indeed the case! I'll edit it in, thank you.

Thanks for thinking about this. I checked Schwede-Shipley "algebras and modules in monoidal model categories", and I believe you're talking about the argument at the beginning of the proof of Lemma 6.2. I don't understand why you can make that assumption on your last sentence, though...

I have a question about your second paragraph (greeting 0-th paragraph not counted). How can you get an $R_E$ which is an R-algebra and a cofibrant R-module (or something equivalent to $R_E$ satisfying these two conditions)?

See also mathoverflow.net/a/195194/6249

I'm in a hurry now so I might have misread, but Proposition 16.2.4 of May-Ponto's "More concise algebraic topology" seems very related.

Proposition IX.2.7 of EKMM (Elmendorf-Kriz-Mandell-May) is related to a particular instance of what you're asking for, I think.

(Above, I meant: if X is an R-module which is positively flat. Couldn't edit it on time). An analogous result is true in EKMM (III.5.1): if R is a cofibrant commutative S-algebra and $X$ is a cell $R$-module, then $\pi$ is a homotopy equivalence of spectra.

@DylanWilson: using different terminology, Proposition 4.2 says: if $f$ is a map in $CAlg(R)$ which is a projective cofibration in $CAlg(R)$, then it is a positive flat cofibration in $CAlg(R)$. This implies that that it is a positive flat cofibration in $Mod(R)$ (so in particular, a flat cofibration in $Mod(R)$, but not necessarily a projective cofibration in $Mod(R)$). As for your last statement: Shipley's Proposition 3.3 proves that an $R$-module which is positive flat, then the standard map $\pi:X^{\wedge_R n}_{h\Sigma_n}\to X^{\wedge_R n}_{\Sigma_n}$ is a stable equivalence.

I'm starting a bounty on the question -- more details on your answer would be very welcome, thanks.

The second is a quote of the paper "Topological Hochschild Homology" by Schwänzl-Vogt-Waldhausen: "We have to distinguish between the associative and commutative case, because the forgetful functor $RCAlg\to RAlg$ does not preserve q-cofibrant objects. This is a well-known phenomenon: in ordinary algebra free associative resolutions use tensor algebras, while free associative and commutative resolutions use symmetric algebras". (Also, thank you for the interest.)

@DavidWhite Two remarks: the first one is that the unit $R\to A$ of a cofibrant $R$-algebra or commutative $R$-algebra is a cofibration of underlying $R$-modules (EKMM, right after VII.4.14).

@DavidWhite $\mathbb S$ is a cofibrant commutative $\mathbb S$-algebra (it is cell), but $\mathbb S$ is not a cofibrant $\mathbb S$-module.