The function $f$ can be written as
$$
f(z)=g(z)+\sum_{k=1}^N\frac{a_k}{z-z_k}
$$
for some $r>1$ and $|z|<r$. Here $g$ is holomorphic and $|z_k|\geq 1$ for $k\in\{1, \dotsc, N\}$. The boundedness of the coefficients follows directly from this decomposition (note that the radius of convergence of $g$ is at least $r>1$).
To finish on the way you started, let $\gamma$ be a circle with centre $0$ and radius $0 < \rho < R$. Then the standard estimate tells you
$$\lvert c_n\rvert = \frac{1}{2\pi}\left\lvert\int_{\lvert w\rvert = \rho} \frac{f(w)}{w^{n+1}}\,dw\right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\rho^n}\,d\varphi \leqslant \rho^{-n}\max \{\lvert f(\zeta)\rvert : \lvert\zeta\rvert = \rho\}.$$
That gives you an upper bound on $\limsup\limits_{n\to\infty} \sqrt[n]{\lvert c_n\rvert}$ in terms of $\rho$. As $0 < \rho < R$ was arbitrary, you get the desired conclusion.
Another way to obtain the result is to use Cauchy's integral formula,
$$f(z) = \int_{\lvert w\rvert = \rho} \frac{f(w)}{w-z}\,dw,$$
for $\lvert z\rvert < \rho$, and expand $\dfrac{1}{w-z}$ into a geometric series, which you know converges uniformly on the circle $\lvert w\rvert= \rho$. Interchanging summation and integration then yields the convergence of the Taylor series in $\lvert z\rvert < \rho$. Let $\rho\to R$.
Best Answer
Yes, this is true. The general result is:
If $f$ cannot be analytically extended outside $U$, then that inequality becomes an equality (since, if $r$ was greater, the power series would be an analytic extension).
The theorem follows directly when looking at the proof of analyticitiy of holomorphic functions. This in turn, is essentially a consequence of the Cauchy integral formula in which you consider contour integrals along a circle around $z$. The point is then realizing that the value of the integral doesn't really depend on the radius of that circle, as long as it is contained in the domain.