Let's denote $D_x = \{ z \in \mathbb{C} \mid |z| < x \} $.
Suppose $f(z)$ is holomorphic in $D_R$ then I want to prove that for any $0 < r < R$ a function $\displaystyle{\frac{f(z+h) - f(z)}{h}}$ uniformly converges to $f'(z)$ on $D_r$.
My attempt goes as follows. Without loss of generality we can assume $|h| < |q| < R - r$. We represent $f(z) = \displaystyle{\sum a_nz^n}$ and then
$f(z+h) - f(z) - hf'(z) = \displaystyle{\sum a_n(z+h)^n - \sum a_nz^n - h \sum na_nz^{n-1}} = h^2 g(z,h)$. Since $|z+h| < |z| + |h| < R$ by our assumption on $h$ we conclude that all series converge in $D_r$. Thus
$$\left| \frac{f(z+h) - f(z)}{h} - f'(z)\right| = |h||g(z,h)|$$
and this is almost it except that I don't know that $g(z,h)$ is bounded. I am sure it is continuous on $\bar D_r \times \bar D_q$ as a function of two variables but it looks like a of work to check this.
So can my approach work or it's easier to do it in some other way?
I'll suggest a different approach. For $z, z+h \in D_R$ is (integrating along the straight segment from $z$ to $z+h$) $$ f(z+h) -f(z) = \int_z^{z+h} f'(w) \, dw = h \int_0^1 f'(z + th) \, dt \, . $$ It follows that $$ \frac{f(z+h) - f(z)}{h} - f'(z) = \int_0^1( f'(z + th) -f'(z))\, dt $$ $f'$ is continuous, and therefore uniformly continuous on any closed disk with radius $r < R$. This shows that the right-hand side converges uniformly to zero for $h \to 0$ on that closed disk.