No, due to a simple reason: Let $C$ be a club class and let $\mu$ be any infinite cardinal (e.g. $\mu = \omega$). Then there is some $\kappa \in C$ such that $\operatorname{cf}(\kappa) = \operatorname{cf}(\mu)$.
Proof. Recursively construct a sequence $(\kappa_{i} \mid i \le \mu)$ be fixing any $\kappa_{0} \in C$, further letting $\kappa_{\alpha+1} := \min C \setminus \\{ \kappa_{i} \mid i \le \alpha\\}$ for successor ordinals $\alpha+1 \le \mu$ and finally, by letting $\kappa_{\lambda} := \sup \\{ \kappa_{i} \mid i < \lambda \\}$ for limit ordinals $\lambda \le \mu$. Now, by construction and the fact that $C$ is closed, $\\{ \kappa_{i} \mid i \le \mu \\} \subseteq C$ and $\kappa_{\mu}$ has cofinality $\operatorname{cf}(\mu)$. Q.E.D.
(The latter follows from the general fact that given a normal sequence $(\kappa_{i} \mid i < \mu)$ of ordinals, we have $\operatorname{cf}(sup_{i < \mu} \kappa_{i}) = \operatorname{cf}(\mu)$.)