Is an $H_0^1$ function continuous to the boundary if it is continuous in the interior?

Not necessarily- let $\Omega = B_1 \cap \{x_3 > 0\}.$ Then $u(x) := (1-|x|^2)\frac{x_3}{|x|}$ is in $H^1_0(\Omega) \cap C^{\infty}(\Omega),$ but $u$ is discontinuous at the origin.

The answer to the follow-up question is negative too. For consider the half-ball $\Omega=\{x\,;\,x_3>0,\,|x|<1\}$. Choose a number $\alpha\in(1,\frac32)$, and a function $\phi\in C^\infty({\mathbb R}^3)$ such that $\phi(x)\equiv1$ for $|x|<\frac13$, while $\phi(x)\equiv0$ for $|x|>\frac23$. Then the function $u(x)=r^{-\alpha}x_3\phi(x)$ belongs to $H^1(\Omega)\cap C(\Omega)$. Its trace, being an element of $H^{1/2}(\partial\Omega)$, is a square-integrable function, hence can be determined by looking away from a negligible Lebesgue set. Thus we look at the trace away from the origin, where $u$ is continuous and vanishes at the boundary. Therefore the trace is $\equiv0$, that is $u\in H^1_0(\Omega)$. Yet, it is not a bounded function.