Are quantum numbers always integers? Quantum numbers are integers for lots of systems. For example, for a particle in an infinite well
$$E_n = E_1 n^2 \text{ with } n = 1, 2, 3\ldots$$
for a quantum harmonic oscillator $$E_n = (n + \frac12) \omega \hbar \text{ with } n = 0, 1, 2, 3\ldots$$
for a quantum rotor $$E_J = \frac{\hbar^2}{2I}J(J + 1)$$ 
and for the hydrogen atom: $$E_n = \frac{E_1}{n^2} \text{ with } n = 1, 2, 3\ldots$$
But in this finite well example, with no exact solutions, there are three bound states for  $v_n = 1.28$, $2.54$ and $3.73$.
The quantum numbers can be seen in a sense as integration constants of the Schrödinger Equation (with the normalisation constant as the second integration constant).
Are the energy quantization numbers always integers or does that depend on the potential function $V(r)$?
 A: No, they are not.
As ACuriousMind pointed out above, quantum numbers are just convention. They may be integers, real numbers, whatever depending on the problem discussed. Note that for example in quantum optics, coherent states $|α\rangle$ are defined in terms of complex quantum numbers, $α\in\mathbb C$. How they translate to actual physical quantities (which they are not) is to be defined in each case.
If turning somebody else's comment into an answer is bad style, please don't downvote, I'll delete the answer then.
Edit: Seems it's not.
A: Quantum mechanics tends to quantize things, forbidding certain "continuous" solutions that would otherwise exist in favor of "discrete" solutions. Whenever those discrete solutions exist, you can assign some integer "quantum numbers" to a system and view solutions as an overlay of those numbers.
A simple answer where they don't exist is if you have the potential $V = 0$, which gives us free-particle wavefunctions in 3D space; you can get normalized Gaussian wavepackets travelling through this potential, but there is nothing to quantize it unless you make the space "wrap around" in some dimension -- so if you try to do $V = 0$ on a torus you might have quantized energies again.
Another one, which mixes them, is $V = \{0\text{ if }|x| > L/2,\text{ else } -V_0\}.$ This has some discrete negative energies of "bound states" but also some "travelling waves" with positive energies; the bound states are likely quantized while the travelling ones are likely continuous. 
Another example occurs in solid-state systems, where often you've got a quantum dot which interacts with phonons and other complicated systems; often these have a simple effect that they effectively "broaden" the discrete energy levels to a Lorentzian distribution in energy. 
A: No, the easiest example is the spin of a fermion (electron, muon, etc...) which is always half integer.
