Consider the union, over all rationals $p/q\in [0,1)$, written in lowest terms, of the line segments starting at $0$, having angle $2\pi p/q$ with the positive real axis, and having length $1/q$. Consider the union of this set with the unit circle and call this set $K$. Then $K$ is a compact, locally connected subset of the Riemann sphere (consisting of the unit circle, the interval [0,1] and countably many "spikes" protruding from zero. Let $U$ be the bounded complementary component of $K$, so $U$ is a simply-connected bounded subset of the plane, with locally connected boundary.
Let $\phi:\mathbb{D}\to U$ be a Riemann map, i.e. a conformal isomorphism between the unit disk and the domain $U$. By the Carathéodory-Torhorst theorem, the map $\phi$ extends continuously to the unit circle. This extension will have uncountably many zeros (one for each "access" to zero from the complement of $K$), but of course the map $\phi$ itself has no zeros.
To get an example of a holomorphic function that has infinitely many zeros, extends continuously to the boundary but has only one zero there (the minimum possible due to continuity) is very easy. For example, restrict the function \sin(z)/z to a horizontal half-strip surrounding the positive real axis, and whose boundary does not pass through any zeros. Precompose with a Riemann map taking the disk to this strip to get the desired map. (This is similar to J.J.'s example as above, but I've divided by z to ensure a continuous extension.)
Edit. It is worth noting that by the F. And M. Riesz theorem, the set of zeros on the boundary has zero one-dimensional Lebesgue measure.
Yes, we can. From root test we know that $\epsilon\cdot\limsup_{n\to\infty}|a_n|^{1/n}\le 1$. Then the radius of convergence of $\sum_{n=0}^\infty a_n(z-x_0)^n$ for complex $z$ is also no less than $\epsilon$, so we can define $F(z)$ as the sum of this series, which is holomorphic on $D$.
Best Answer
No, this is in general not possible. Take e.g., $$ f(z) = \sum_{n=1}^\infty \frac{z^{n!}}{n!} $$ This series converges absolutely and uniformly for $|z| \le 1$, with $$ |f(z)| \le \sum_{n=1}^\infty \frac{1}{n!} = e-1 $$ but the derivative $$ f'(z) = \sum_{n=1}^\infty z^{n!-1} $$ does not extend holomorphically to any domain strictly containing the open unit disk. (The radial limit at any root of unity is infinite.)