What you suggest should work. We're going to transport the model structure on the category of chain complexes via the adjunction using the criteria given in this answer: http://mathoverflow.net/questions/92538/transporting-model-structures-via-adjunctions
(A proof of this transport theorem for our case, where adjunction functor is monadic, can be found, for example, as Lemma 2.3 in this paper of Schwede and Shipley, http://homepages.math.uic.edu/~bshipley/monoidal.pdf)
We have an adjunction between the functors
$$\text{Free}: \mathbf{Ch}_{\mathcal{O}} \rightarrow \mathcal{O}\text{-}\mathbf{dga}$$ and
$$\text{Forget}: \mathcal{O}\text{-}\mathbf{dga} \rightarrow \mathbf{Ch}_{\mathcal{O}}$$
where the first is left adjoint to the second. Since $\mathcal{O}$-modules form a Grothendieck abelian category, there is a combinatorial model structure on $\mathbf{Ch}_{\mathcal{O}}$ with the fibrations and weak equivalences you described. Every object in both categories is small, since they are presentable.
Now I need to show that everything that can be obtained by sequential limits from cobase changing the arrows $\text{Free}(g)$,where $g$ is a (generating) acyclic cofibration in $\mathbf{Ch}_{\mathcal{O}}$, is a quasi-isomorphism. I think this is true, but I'm not sure so I'll include my argument in case there's something wrong with it.
First I claim that the relative tensor product $\text{Free}(D) \otimes_{\text{Free}(C)} (-)$ is exact on chain complexes. Indeed, given an exact sequence of chain complexes $0 \rightarrow X \rightarrow Y \rightarrow Z \rightarrow 0$ I can form the diagram

All three columns and the top two rows are exact, since free algebras are free as graded modules and the exactness diagrams of chain complexes is determined by exactness as graded objects. Therefore the bottom row is exact (diagram chase or spectral sequence argument.) (The unadorned tensors are over $\mathcal{O}$).
Now I claim that cobase changing a morphism $\text{Free}(C) \rightarrow \text{Free}(D)$ where $C \rightarrow D$ is a monomorphism that's a quasi-isomorphism, gives a quasi-isomorphism. Indeed, we have a convergent spectral sequence
$$
\text{Tor}_{p,q}^{H^*F(C)} (H^*F(D), H^*A) \Rightarrow H(F(D) \otimes^{\mathbb{L}}A)
$$
But since $F(D)$ is a flat $F(C)$ module this converges to the cohomology of $F(D) \otimes_{F(C)} A$. On the other hand,
$$
H^*F(C) \rightarrow H^*F(D)
$$
is an isomorphism so the $E_2$-term collapses to an edge, and moreover the edge is just $H^*A$. The edge homomorphism is then an isomorphism, but the edge homomorphism is precisely the map induced by $A {\rightarrow} F(D) \otimes_{F(C)} A$, whence this map is a quasi-isomorphism, which was to be shown.
Since sequential colimits in the category of algebras are the same as those in the category of chain complexes, we already know that sequential colimits of quasi-isomorphisms are quasi-isomorphisms. The result follows.
(I didn't use that the arrow $F(C) \rightarrow F(D)$ was monic... so that worries me.)