The answer is "Quite often yes, but the error terms are seldom as good as in the meromorphic case". The reason the asymptotics for the meromorphic functions works is that we know the exact coefficients for $c_k(z-z_0)^{-k}$ and can always remove the singularity by subtracting a few terms of this kind thus increasing the radius of convergence and boosting the decay of the coefficients, so that the error is exponentially small compared to the main terms.

When you have some branching singularity $S(z)$ (like $\log(z-z_ 0)$ or $\sqrt[]{z-z_ 0}$) for which you know the coefficients when it stands alone, you can still play the trick to get the asymptotics for a function $F(z)S(z)$ where $F$ is alnalytic in some larger disk by writing the sum as $F(z_ 0)S(z)+(F(z)-F(z_ 0))S(z)$ and noting that the second term is smoother than the first one on the boundary circle, so its Fourier coefficients decay a bit faster. If you take more terms in the Taylor series for $F$ at $z_ 0$, you can get any degree of smoothness in the remainder you wish, so the error terms can be made of size $n^{-k}$ times the leading terms with any given $k$, but that's about as far as you can go unless you want integral expressions instead of algebraic quantities.

In the latter case, you should make a radial cut starting at singularity and ending at some bigger circle and take the contour integral over that larger circle and 2 sides of the cut in the Cauchy formula. The integral over the larger circle will decay exponentially faster than the typical coefficient, so all asymptotics will come from the cut. The point is that the integral over the cut is not that of something oscillating but that of a fast decaying function, so you can use real variable methods to find its asymptotics in nice algebraic terms (or you can just leave it as is).