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.
Let me summarize the comments: I'm assuming that we talk about meromorphic functions.
As long as we don't control the singularity at infinity, poles and zeroes and their multiplicities are far from enough to determine the function.
Indeed, if $h:\mathbb{C} \to \mathbb{C}$ is any entire function then $\exp{h}$ has no zeroes and no poles and thus $f(z)$ and $g(z) = f(z) \cdot \exp{(h(z))}$ have the same poles and zeroes.
This is the “only” ambiguity: if $f(z)$ and $g(z)$ are meromorphic functions with the same set of poles and zeroes and with the same multiplicities then $k(z) = \frac{f(z)}{g(z)}$ only has removable singularities and thus it extends to an entire function: $k : \mathbb{C} \to \mathbb{C}$. It is not hard to show that $k$ has no zeroes, and a standard fact from complex analysis tells us that a function $k: \mathbb{C} \to \mathbb{C} \smallsetminus \{0\}$ can be written as $k(z) = \exp{(h(z))}$ for some entire function $h$. (a)
There is the surprising fact that for every closed discrete subset $D \subset \mathbb{C}$ (which may well be infinite) and any assignment $o: D \to \mathbb{Z}$ there exists a meromorphic function $f: \mathbb{C} \to \mathbb{D}$ such that $f$ has a zero of order $k$ at $z_0$ if and only if $z_0 \in D$ and $o(z_0) = k \gt 0$, and a pole of order $k$ at $z_0$ if and only if $z_0 \in D$ and $k = -o(z) \gt 0$. Of course, if $D$ is finite, this is easy to achieve (and a good exercise to do), but if $D$ is infinite you need to work a bit. This is Weierstrass's factorization theorem.
My favorite reference is Remmert's Classical topics in complex function theory, see chapter 3, p.73ff and also chapter 4, p.89ff for more on this.
Finally, if we do control the singularity of the meromorphic function $f$ at infinity and decide that it should be non-essential (that is to say $f(1/z)$ has either a pole or a removable singularity at $0$), then it follows that $f$ is a rational function.
This is proved in detail e.g. in Theorem 4.7.7 on page 144 of Greene-Krantz, Function theory of one complex variable.
(a) Indeed, $h(z)$ can be chosen to be
$$
h(z) = C \cdot \int_{1}^z \frac{k'(w)}{k(w)}\,dz
$$
where $C$ is such that $e^C = k(1)$ and a straightforward computation shows that $k(z) \cdot \exp{(-h(z))}$ has zero derivative while $k(1)e^{-h(1)} = 1$ so that $k(z) = \exp{(h(z))}$ as desired (note that $h$ is holomorphic and well-defined precisely because $k$ has no poles and zeroes).
Best Answer
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.