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.
The simplest examples are probably infinite Blaschke products:
$$f(z) = \prod_{n=1}^\infty \frac{z-z_n}{1-\bar z_n z}.$$
You can check that this product converges when your condition (which is often called the Blaschke condition) is satsified.
Best Answer
Let $f(z)=\sum_{n=0}^{\infty}a_nz^n$ be the series expansion of $f$ at zero, and suppose that this series has radius of convergence $R$. Then it's a general fact about holomorphic functions that $f$ must have a singularity on the circle $|z|=R$ (for a proof, see Big Rudin Chapter 16).
Since by assumption $f$ is continuous on $|z|=1$, this means that $R>1$, so we can still apply the identity theorem to conclude that $f\equiv 0$.
Edit: It appears I mis-remembered Rudin's definition of singularity here. He defines a singular point to be a boundary point which does not admit an analytic continuation. So just continuity on the boundary is not enough for my argument above.
For a construction relevant to your question, see this answer: Zeros of a holomorphic function on the boundary of a closed region