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.
If $f$ is continuous on $\overline{\mathbb D}$ then it is bounded. In that case there is a strong consequence of Jensen's formula. If $|f| \leq M$, $f(0) \neq 0$ and $S$ is the (possibly infinite) multiset of roots of $f$ on $\mathbb D$ then
$$\prod_{z \in S} |z| \geq \frac{|f(0)|}{M} > 0.$$
This implies that $$\sum_{z \in S}(1-|z|) < \infty.$$
So, loosely speaking, the roots of $f$ must quickly converge to the unit circle. The condition $f(0) \neq 0$ can be circumvented. There is a unique integer $m \geq 0$ such that $$g(z) = \frac{f(z)}{z^m}$$ is holomorphic and $g(0) \neq 0$. If $S$ is the multiset of non-zero roots of $f$ then the inequality above becomes
$$\prod_{z \in S} |z| \geq \frac{|f^{(m)}(0)|}{m! \, M}.$$
An explicit example of a function $f$ that is holomorphic on $\mathbb D$, continuous on $\overline{\mathbb D}$ and has infinitely many roots is
$$ f(z) = (1-z^2) \sin(\operatorname{atanh}(z)).$$
To understand this example, note that $\operatorname{atanh}$ maps $\mathbb D$ biholomorphically onto the strip $\{ z \mid -\frac{\pi}{4} < \operatorname{Im}(z) < \frac{\pi}{4} \}$ where $\pm 1$ maps to $\pm \infty$. Now note that $\sin$ is bounded and has infinitely many roots on this strip. The factor $1-z^2$ makes sure that it is continuous at $\pm 1$ (and has roots there as it should).
Best Answer
It is obviously false if the domain is not connected: Take $f(z)=1$ for $|z|<2$ and $f(z)=0$ for $|z-10|<1$.
For the proof when the domain is connected consider $g(z)=e^{if(z)}$ and note that $|g(z)|=1$ for all $z$ with $|z|=1$. Do you know how to show that $g$ is a constant? [This has been proved many times on this site].