I don't think such a description is possible:
The fields in question ought to be exactly the residue fields of points in $X=Spec(R)$ (It's enough to consider affine schems here). Clearly any point $x\in X$ has a residue field $K_x$, which gives us a morphism. $Spec(K_x)\to X$.
Similarly, for any morphism $f: R\to K$, $K$ any field, consider $Ker(f)$. Since $Im(f)$ is a subring of $K$, it is an integral domain. Thus $Ker(f)$ is prime $p$. Thus this morphism determines a point of $X$. This point has residue field a subfield $K'$ of $K$, containing $im(f)$. So we can think of $p$ being determined by the induced map $f: R\to K'$. (The associated map $R_p\to K'$ will be surjective with kernel $p$).
So geometric points correspond to $\it{extensions}$ of residue fields of points in $X$, so there will be many geometric points which determine one regular point.
The subtlety is that if $K_x$ is a residue field at $x$, and $K_y$ is a residue field at $y$, we could have $K_x$ an extension of $K_y$. Thus a map $R\to K_x$ whose image lies in $K_y$ actually corresponds to $y$, when $y$ is already determined by the restriction of this map. So a functor of points description shouldn't be possible.
As an aside, closed points are certainly exactly surjective ring maps $R\to K$, and points in general will be exactly the surjective maps from $R_p\to K$, for $p$ a prime, which isn't very helpful.