This should be possible to solve in the same way we do the ordinary catenary problem, by variational calculus.
Suppose the angular separation between the endpoints is $\Delta$, where we could define $\Delta = \frac{D}{R}$ if I understand the problem correctly. Let the shape of the rope be given by a function $r(\phi)$, and write the potential energy of the rope as
$$U_G = \int_0^\Delta \frac{GM}{r}\lambda\sqrt{r'^2 + r^2}\mathrm{d}\phi$$
Here $\lambda$ is the linear mass density of the rope, and I've left the $\phi$ dependence of $r(\phi)$ implicit.
We also need a constraint that enforces the total length of the rope remain fixed at $L$. According to the method of Lagrange multipliers, we'll have to add a term
$$U_L = \int_0^\Delta l\sqrt{r'^2 + r^2}\mathrm{d}\phi$$
for a total of
$$U = \int_0^\Delta \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r'^2 + r^2}\mathrm{d}\phi$$
If we define $T$ to be the integrand,
$$T(r, r', \phi) = \biggl(\frac{GM}{r}\lambda + l\biggr)\sqrt{r'^2 + r^2}$$
this can be solved by the Euler-Lagrange equation,
$$\frac{\mathrm{d}}{\mathrm{d}\phi}\frac{\partial T}{\partial r'} = \frac{\partial T}{\partial r}$$
which gives
$$(r''r - r'^2)(GM\lambda + lr) = lr(r'^2 + r^2)$$
There does appear to be an analytic solution to this equation, but Mathematica makes it look rather icky to say the least... I'll have to come back to this and try to find some sensible solution and/or physical interpretation since I'm out of time right now.