In Theorem 6.14 of Gilbarg and Trudinger, it is in fact assumed that $\partial \Omega$ is of class $C^{2, \alpha}$. They discuss a variety of generalizations, but not of the global $C^{2, \alpha}$ estimate; rather, they generalize the existence theory implied by it.
The condition that $\partial \Omega$ is of class $C^{2, \alpha}$ is sort of sharp. Indeed, take a domain $\Omega$ which is, say, at least $C^{1, \alpha}$, with $y \in \partial \Omega$, and find a harmonic function in a neighborhood of $y$ which vanishes along $\partial \Omega$ and is positive on $\Omega$ (this is easy by solving some Dirichlet problem, for example). By basic regularity theory, $u \in C^{1, \alpha}(\bar{\Omega})$ locally, and by Hopf lemma the inward normal derivative of $u$ is strictly positive along $\partial \Omega$.
Let $\partial \Omega$ be locally given as a graph $\{x_n = g(x')\}$ with $\nabla' g(y') = 0$ for our designated point $y$ (and $\Omega = \{x_n > g(x')\}$ locally). Then we can take the equation $u(x', g(x')) = 0$ and differentiate it in the $x'$ variables, to get $\nabla' u(x) + \nabla' g(x') \partial_n u(x) = 0$. This gives the formula
$$
\nabla' g(x') = \frac{\nabla' u(x)}{\partial_n u(x)}
$$
The denominator is nonzero near $y$, as at $y$ itself we have $\partial_n u$ is the inward normal derivative (so: nonzero) and $\nabla' u(y) = 0$ as $\nabla'g(y') = 0$, while at nearby points it follows from the continuity of $\nabla u$.
But then if $u \in C^{2, \alpha}(\bar{\Omega}\cap B_r(y))$, this relation implies that $g$ is also $C^{2, \alpha}$. So we can conclude: (1) If a $C^{1, \alpha}$ domain has $\partial \Omega$ not $C^{2, \alpha}$ at a point, then there exists a harmonic function which is not $C^{2, \alpha}(\bar{\Omega})$ locally. This can be strengthened to say that any solution to the Poisson equation (or any equation! this isn't a PDE argument, this is the implicit function theorem) with nonzero normal at this point will fail to be $C^{2, \alpha}$ up to the boundary.
You can ask further questions here. For example, you can ask, well, if the normal derivative of $u$ does vanish at a point, is $u$ more regular there? There are theorems in that direction in the unique continuation literature. You can ask, what if the domain is not $C^{1, \alpha}$? Here it is useful to understand what happens on cones in the plane (i.e. consider the harmonic functions $\Re(z^\alpha)$ for $\alpha \geq 1/2$ on regions between two rays where they vanish): for cones with opening less than $\pi/2$, the harmonic function actually vanishes fast enough at the origin that it is $C^{2, \alpha}$ for some $\alpha$; for larger angles it is not $C^{2, \alpha}$ (except the special case of opening $\pi$). So you can say something if all of your corners are very acute.
For your second question, can you say that $u \in C^{2, \alpha}(\bar{\Omega}\setminus \{\text{bad set}\})$: sure. Use local boundary estimates, like Cor. 6.7 in Gilbarg-Trudinger. Everything in this theory is local.
Best Answer
To have unique solvability for the Dirichlet problem you can assume a lot less regularity on the boundary, but then you won't necessarily have that $u$ is $C^{2,\alpha}$ up to the boundary. For example, in Elliptic Partial Differential Equations of Second Order by Gilbarg and Trudinger, we have:
In fact, they claim that the assumption that $\Omega$ satisfies an exterior sphere condition at every boundary point can be relaxed to only satisfying the exterior cone condition (they leave it has an exercise to prove this). Also, in the following theorem (Theorem 6.14), they prove what you suspected, that is, if $\Omega$ is $C^{2,\alpha}$ and $\varphi \in C^{2,\alpha}(\overline\Omega)$ then there is a unique solution $u\in C^{2,\alpha}(\overline\Omega)$.