I think you are struggling with the differences between implicit and explicit time dependence, which was well explained here.
It is true that, since $r=r(t)$ and generally $g=g(r)$, we think about $g$ varying on time, but this variation is implicit. Explicitly, we could say that $g\neq g(t)$ and
$$
{\partial g\over \partial t} = 0.
$$
The total time derivative, on the other hand, would be not zero, since
$$
{dg\over dt} = {dr\over dt}{dg\over dr}
$$
And the two quantities on the r.h.s. are generally not zero. We have
$$
g(r) = {GM\over r^2} \rightarrow {dg\over dr} = {-2GM\over r^3} = -\frac 2r g(r)
$$
So we can write
$$
{dg\over dt} = -\frac 2r {dr\over dt} g(r)
$$
There is no scape but to calculate $r(t)$ explicitly and get the derivative, which people already explained in the comments that it is a hard task. What you can do is to consider some approximations for $r(t)$, but I'm not sure how much insightful it can be, since at the end you only have an implicit time dependence. In fact, the expression above will only give you $g(r(t))$ again, and it could only be thought as a explicit function of time if we substitute $r$ by $r(t)$ in the expression of $g$ and write $g(t) \equiv g(r(t))$.
Consider for example that for some reason, a body not following a Keplerian orbit have constant negative radial velocity ${dr\over dt} = - C$ for $C>0$ (it is going to the center of earth). Then $r(t) = r_0 - Ct$ and we have
$$
{dg\over dt} = {2C\over r_0 - Ct} g \rightarrow \ln g = -2\ln(r_0-Ct) + K\rightarrow g(t) = {K\over (r_0 -Ct)^2}
$$
Which is the same result we get by inserting $r(t)$ directly in the expression of $g(r)$ and setting the constant $K=GM$.
