Can you determine from the minors if the presented module is free?

Even for $A$ noetherian it's not true that $M$ is free.

Proposition 1.4.10 from Bruns and Herzog, Cohen-Macaulay Rings, says the following:

Let $A$ be a noetherian ring and $M$ a finitely generated $A$-module with a finite free presentation $F_1\stackrel{\varphi}\to F_0\to M\to 0$. TFAE:

(i) $I_s(\varphi)=R$ and $I_{s+1}(\varphi)=0$;

(ii) $M$ is projective and $\operatorname{rank}M=\operatorname{rank}F_0-s$. (Here $I_t(\varphi)$ denotes the ideal generated by the $t\times t$ minors of the matrix $\varphi$.)

This shows that in your case $M$ is projective of finite rank $m-s$.

No, it is not true that locally free implies free on an affine scheme. Here are some details.

Consider a noetherian ring $A$ and the corresponding affine scheme $X=\text {Spec(A)}$.
A finitely generated module $M$ over $A$ corresponds to a coherent sheaf $\tilde M=\mathcal F$ on $X$.
We have the equivalences $$M \;\text {is projective}\iff \mathcal F \;\text {is locally free}\quad (I)$$ and $$M \;\text {is free}\iff \mathcal F \;\text {is free} \quad (II)$$

However the conditions $(II)$ are much stronger than $(I)$:

If the rank of $M$ (or equivalently of $\mathcal F$) is one, for example, the corresponding projective modules constitute the Picard group $\text {Pic}(A)$, denoted $\text {Pic}(X)$ in the geometric setting.
So every non-zero element of $\text {Pic}(A)$ yields a non-free locally free coherent sheaf of rank one.
But do there exist rings with $\text {Pic} (A)\neq 0$ ?
Sure: I could give you examples but I prefer to impress you by stating Claborn's unbelievable (but true!) result:

Given an arbitrary abelian group $G$, there exists a Dedekind domain $A$ with $\text {Pic}(A)=G.$

Edit on your "last thought"
Indeed Quillen-Suslin is very relevant: they proved that on $\mathbb A^n_k$ ($k$ a field) , corresponding to $A=k[T_1,\ldots, t_n]$, $(I)$ and $(II)$ are equivalent: every locally free sheaf on affine space is free .
This is analogous to the result that every topological vector bundle is trivial on $\mathbb R^n$, but much more difficult.
(And as a parenthetical remark, it is a crying shame that Suslin, one of the greatest living algebraists/algebraic geometers never got a Fields medal.)