NOTE: I assumed that the question was about real valued functions. (The OP did not specify this and this seemed to me as the most probable explanation.)
Not necessarily.
Note that if $f: \omega \to \mathbb R$ is any sequence convergent to $0$, then the continuous extension $\overline f:\beta\omega \to \mathbb R$ fulfills $\overline f(x)=0$ for every free ultrafilter (=for every point $x\in \beta\omega\setminus\omega$).
There are probably many ways how you can see that the above holds. (Depending on your favorite definition of $\beta\omega$, some of them might be clearer for you then others.) For instance, you can notice that $\overline f(x)$ is a cluster point of the sequence $(f(n))$ and in the case of convergent sequence, there is only one cluster point.
Thus choosing $f(n)=\frac1n$ gives an easy counterexample.
(Note that I am identifying integers with principal ultrafilters, which is quite usual in this context.)
About your edited question:
Are there any first-countable compact Hausdorff spaces or at least compact Hausdorff spaces with a point $x_0$ having a countable nbhd basis such that any real continuous function from this space with $f(x_0)=0$ vanishes at some isolated point distinct with $x_0$?
Every compact Hausdorff space is is normal. In a normal space, every closed $G_\delta$-set is zero-set, i.e. it is equal to $f^{-1}(0)$ for some real-valued function. This is a consequence of Urysohn's lemma. See e.g. Theorem 4 in Henno Brandsma's Useful theorems on normal spaces.
Now $\{x_0\}$ is a closed set and, since you also assume that it has a countable neighborhood basis, it is also a $G_\delta$-set. So there exists a real valued function on $X$ such that $f^{-1}(0)=\{x_0\}$.