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.
As explained in comments by Daniel Fischer, this cannot happen. Here's a slightly more general version, with $\mathbb C$ replaced by an arbitrary domain $\Omega\subset \mathbb C$.
We can write $\Omega$ as the union of a countable family of compact sets $K_n$, for example by letting $K_n=\{z\in \Omega: \operatorname{dist}(z,\partial \Omega) \ge 1/n, \ |z|\le n\}$.
Since the zero set $Z$ has no limit points, each intersection $Z\cap K_n$ is finite. Hence
$$Z = \bigcup_n (Z\cap K_n)$$
is countable.
Best Answer
As others have said, it comes down to the nonexistence of an uncountable discrete subset $Z$ of an open subset $U$ of the complex plane $\mathbb{C}$. The point of this answer is to record a proof of this that I find simplest. Namely, the space $\mathbb{C}$ is second countable: there is a countable base for the topology (take, e.g., open balls with rational radii centered at points $x+yi$ with $x,y \in \mathbb{Q})$). But every subspace of a second countable space is second countable: just restrict a countable base to the subspace. In particular $U$ is second countable and so is the putative uncountable discrete subset $Z$ of $U$. But this is absurd: since the only base of a discrete space consists of all the singleton sets, an uncountable discrete space is not second-countable!