When do the eigenkets of an operator form a(n) (orthogonal) basis for the Hilbert space? When do the eigenkets of an operator form a(n) (orthogonal) basis for the Hilbert space? Is it always the case when the operator is Hermitian? Does the operator need to be Hermitian?
 A: Restricting to finite-dimensional Hilbert spaces, you have a complete basis of orthonormal eigenkets if and only if the operator is normal, meaning that it commutes with its adjoint, $[A, A^\dagger] = 0$.
In particular, Hermitian operators are normal since they are equal to their adjoints, anti-Hermitian operators are normal, and unitary operators are normal since $[A, A^\dagger] = [A, A^{-1}] = 0$. That covers most options you see in quantum mechanics classes.
In the infinite-dimensional case, there are many more subtleties, which I honestly don't have the slightest understanding of. However, I've never seen a realistic setup where the subtleties cause one to get the wrong physical answer, as long as you follow your nose.
A: 
Does the operator need to be Hermitian?

No, the operator doesn't need to be Hermitian. Here is an example.

Is it always the case when the operator is Hermitian?

No, the operator also needs to be compact. In the finite case, however, you don't need this extra hypothesis.

When do the eigenkets of an operator form a basis for the Hilbert space?
When is this basis orthogonal?

In general, the Spectral Theorem (at least, the version useful to the study of Hilbert spaces in quantum mechanics) tells us that if an operator is compact and normal, its eigenkets will form an orthonormal basis for the Hilbert space (see Kowalski's Spectral Theory, p. 20). I don't know if one could find a different version of the Spectral Theorem, with weaker hypotheses, that doesn't give you orthonormality, but I can't see why that would be useful.
