The answer is yes.

There can be $f\in W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega)$ such that $f=0$ on $\partial \Omega$ but $f\notin W^{1,p}_{0}(\Omega)$.

If $f\notin C({\bar{\Omega}})$, then its trace on $\partial \Omega$ can be different from zero, and in that case $f\notin W^{1,p}_{0}(\Omega)$.

Example: simply the characteristic function $\chi_{\Omega}$

The answer is No.

There can be $f\in W^{1,p}(\Omega)\cap C(\Omega) \cap L^{\infty}(\Omega)$ such that $f=0$ on $\partial \Omega$ but $f\notin W^{1,p}_{0}(\Omega)$.

If $f\notin C({\bar{\Omega}})$, then its trace on $\partial \Omega$ can be different from zero, and in that case $f\notin W^{1,p}_{0}(\Omega)$.

Example: simply the characteristic function $\chi_{\Omega}$

The trace of a function $f \in C(\Omega)$ at $x\in \partial \Omega$ is the limit of $f$ when we move from the interior of $\Omega$ toward $x$. If $f\in C(\Omega) \cap L^{\infty}(\Omega)$ then its trace exists. If $f\in C({\bar{\Omega}})$ then the trace (the limit) at any $x\in\partial \Omega$ coincides with $f(x)$. So, assuming $f=0$ on $\partial \Omega$ and continuity on $\Omega$ are not enough.