What form of explicit description do you have in mind?
If you let $n$ vary, any finitely generated field of extension of $k$ can occur as an $A(\mathfrak p)$,
as is easily seen, and so classifying the possible $A(\mathfrak p)$ is the same as classifying the possible f.g. field extensions of $k$, which is essentially the same thing as classifying algebraic varieties over $k$ up to birational equivalence, a fairly difficult problem.
One thing you can say is that the transcendence degree of $A(\mathfrak p)$ over $k$ is the equal to the Krull dimension of $A/\mathfrak p$, which in turn is equal to $n - \mathrm{height} \, \mathfrak p$.
Let's consider the case $n = 2$, as an example. Then $\mathfrak p$ is either maximal
(equivalently, $A(\mathfrak p) = k$), the zero ideal (equivalently,
$A(\mathfrak p) = k(x_1,x_2)$), or height one (equivalently, $\mathfrak p$ is principal). In the height one case, the field $A(\mathfrak p)$ is the function field of an algebraic curve over $k$.
Any f.g. field extension of $k$ of transcendence dim'n $1$ can arise as
an $A(\mathfrak p)$, and for a fixed such extension $K$, the classification of the
possible $\mathfrak p$ for which $A(\mathfrak p) \cong K$ is equivalent
to the classification of the (possibly singular) plane models of the curve $C$ corresponding to $K$.
The classifiation of the different possible $K$ is the problem of classifying all curves over $k$, which is an elaborate theory (involving the concept of genus of a curve, the moduli space of curves of a given genus, and so on).
Just to be even more specific, if $\mathfrak p = (f)$ with $f$ of degree one or two, then $A(\mathfrak p) \cong k(x).$ But in fact there are irreducible $f$ of every degree for which $A\bigl((f)\bigr) \cong k(x)$, although for most $f$ of degree $> 2$, the field $A\bigl( (f)\bigr)$
will not be isomorphic to $k(x)$. (E.g. if $f$ is of degree three, then the condition for $A\bigl((f)\bigr)$ to be isomorphic to $k(x)$ is that the homogenization of $f$ describe a singular cubic.)
Let $R, M$ be as in the answer of Andrew. Let $n$ be the (constant) rank of $R_\mathfrak p/\mathfrak pR_\mathfrak p$ for all prime ideals $\mathfrak p$ of $R$.
Fix any $\mathfrak p$. By Nakayama's lemma, $M_\mathfrak p$ is generated by $n$ elements $x_1,\cdots, x_n\in M$. Writing the elements of a generating system of $M$ as linear combinations of the $x_i$'s with coefficients in $R_\mathfrak p$, we find an $f\in R\setminus \mathfrak p$ such that $M_f$ is generated by $x_1,\cdots, x_n$. This implies that we can cover $\mathrm{Spec}(R)$ by various Zariski open subsets $D(f)$ such that $M_f$ is generated by $n$ elements.
Let's show that $M_f$ is free of rank $n$. For simplicity, we replace $R, M$ by $R_f, M_f$. Let $R^n\to M$ be a surjective linear map with kernel $N$. For any $\mathfrak p$, the induced map
$$(R_\mathfrak p/\mathfrak pR_\mathfrak p)^n\to M_\mathfrak p/\mathfrak pM_\mathfrak p$$
is surjective, hence bijective. This implies that $N_\mathfrak p\subseteq (\mathfrak p R_\mathfrak p)^n$ and $N\subseteq (\mathfrak p )^n$. By Krull's intersection theorem, and because $R$ is reduced, we get $N=0$ and $R^n\simeq M$.
In particular, $M$ is a flat and finite generated module over $R$. When $R$ is noetherian, it is known thant $M$ is then projective. Otherwise the same result holds if $M$ is finitely presented.
To return to your case where $R$ is a finitely generated algebra over an algebraically closed field $F$, the methods works with the following two observations: (1) $M\otimes_R R/\mathfrak m=M_\mathfrak m/\mathfrak mR_{\mathfrak m}$ and $R/\mathfrak m=F$; (2) By Hilbert's Nullstellensatz, the intersection of all maximal ideals is $0$ (because $R$ is reduced).
Best Answer
No. If $M$ is f.g. over $A$ then $M\otimes_A (A/\mathfrak m)$ will be finite dimensional over $F$ for every maximal ideal $\mathfrak m$ of $M$ (since $A/\mathfrak m = F$, and change of scalars preserves the property of being finite dimensional).
So your question amounts to asking whether every f.g. module over $A$ that is infinite dim'l over $F$ is free, and the answer is no.
E.g. Suppose that $n = 1$, and take $M = A \oplus F,$ with $x$ and $y$ acting via $0$ on the second summand.
E.g. Suppose that $n = 2$, so $A = F[x,y].$ Let $M = F[x]$, regarded as an $A$-module by having $y$ act via $0$.
E.g. Suppose that $ n =2$, and let $M = (x,y) \subset F[x,y]$.