The confusion between your desired statement and the statement from the nlab is arising because of the confusion between chain complexes and cochain complexes.

The answer to your first questions is "yes." I'm not sure about the second question, but I would guess the answer is "no."

Because your fibrations are surjections in positive degrees, you are working with chain complexes, where the differential reduces degree. On the nlab link, you can see that they specify that they are working with cochain complexes, where the differential goes up in degree, and that their fibrations are surjections in all degrees. The underlying categories, while superficially similar, are quite different, and the model category structures make this more so.

Your model category structure comes from transferring the model category structure on non-negatively graded chain complexes, which is cofibrantly generated. For $i\ge 0$, let $S^i$ denote the chain complex which is $k$ in degree $i$ and let $D^{i+1}$ denote the acyclic chain complex which is $k$ in degree $i$ and $i+1$. Then the cofibrant generators for chain complexes are $J$, the inclusions of $0$ into $D^{i+1}$ (for acyclic cofibrations) and $I$, the inclusions of $S^i$ into $D^{i+1}$ (for cofibrations), along with the extra weird generator $0\to S^0$ (which we could think of as $S^{-1}\to D^0$, I guess). Then applying a free functor to these examples yields the generating cofibrations for the model structure on algebras or commutative algebras.

Then a $(\mathtt{C})\mathtt{DGA}^+$-map is a cofibration precisely if it is a retract of a relative $F(I)$-cell complex. I'm not so good at messing around with cardinals but it's certainly clear that pushing out along a map in $F(I)$ is an almost free extension. So unless there are problems coming from some large cardinal (I'm pretty sure there are not) you get that relative $F(I)$-cell complexes are (some subset of) almost free extensions. But what subset?

The point is that if this is going to be a cell complex, you've got to attach your cells in an order, and you can only glue cells onto previously attached cells. This is where the difference between chain complexes and cochain complexes really comes into play. In chain complexes, you can choose to attach your cells inductively, starting with the zero cells from the weird generator and then proceeding up the tower, attaching all $n$-cells in any arbitrary order. This reordering of your cells can't mess anything up because adding $n$-cells can never touch anything above $n$. This gives you an automatic filtration. Your $0$-cells are never attached to anything; your $1$-cells can only be attached to $0$-cells, your $2$-cells can only be attached to things you built out of $0$-cells and $1$-cells, and so on. So in chain complexes, every almost free extension is automatically Sullivan.

This is definitely not true in cochain complexes, where you can find Borromean type examples ($\Lambda(x,y,z)$, all generators in degree $1$, with the differential $dx=yz, dy=zx, dz=xy$). This happens because the differential from degree one can interact with generators from degree $1$.

This is tagged "reference request" but I don't actually know a clean reference; most of the careful expositions are done with unbounded complexes and the few I know that use non-negatively graded complexes seem not to try to characterize the cofibrations.

For the second question, you would somehow need to be able to see that the inclusions $S^i\to D^{i+1}$ are all cofibrations; I don't see how to do that starting only with the data of the cofibrant objects. All I know how to do is build things with pushouts and sequential colimits and so on. In that situation I don't see any way to obtain cofibrations $X\to Y\cong_{\text{linearly}}X\oplus Z$ such that there is a differential that goes from $Z$ to $X$. I'd love to be wrong about this!