The answer is no, this is never an ultrafilter. This uses choice. We assume that the cofinality of $\kappa$ is bigger than $\omega$, else, $C$ has a subset of type $\omega$ cofinal (which is automatically club), and this can be split into two disjoint cofinal (therefore club) sets.
If the cofinality of $\kappa$ is larger than $\omega_1$, simply note that for any regular $\alpha<\kappa$, $\{\beta<\kappa:\beta$ has cofinality $\alpha\}$ is stationary, i.e., it meets every club (consider the $\alpha$-th member of the increasing enumeration of the club. These sets are disjoint for different $\alpha$, and if $\kappa>\omega_1$, there are at least two regular $\alpha$ below $\kappa$, namely $\omega$ and $\omega_1$. This shows that we can split $C$ into disjoint stationary sets (and therefore, neither is club, since a stat. set must meet every club).
All I've used so far is that $\omega_1$ is regular.
To show that clubs on ordinals of cofinality $\omega_1$ can be split into disjoint stationary sets requires more choice. In fact, it is consistent that the club filter on $\omega_1$ is an ultrafilter. This is the case, for example, in models where the axiom of determinacy holds. (But it is strictly weaker than determinacy. A measurable suffices in consistency strength.)
The typical argument from choice uses the existence of Ulam matrices. See definition 12 and the following paragraph in this blog entry. An $\omega\times\omega_1$ Ulam matrix gives us that every stat. subset of $\omega_1$, in particular every club, can be split into $\omega_1$ disjoint stat. sets. So the club filter is not an ultrafilter, and we are done.
Ulam's argument also shows that any stat. subset of $\kappa^+$ can be split into $\kappa^+$ stat. sets, and Solovay proved the same for any reglar $\lambda$, not just a successor. This, of course, gives the result that no club filter is an ultrafilter, but I wanted to point out that only in cofinality $\omega_1$ we need that much choice to carry out the argument.