Can a magnet rip protons from a nucleus? Pretty self explanatory. I’m wondering if the strong nuclear force could be overcome by a strong enough magnet?
 A: Maybe.   The action of a magnetic field would change the nuclear orbitals if it
were strong enough, and would determine a new region of stability for the 
nucleus.   A 'destabilized' nucleus could then decay in a variety of
ways, including spontaneous fission.
Well, theoretically, it could.   The magnetic field requirements for any
such effect are EXTREMELY high, and any apparatus capable of generating
such a field would also have its atomic (electron) orbitals disturbed
so as to destabilize any and all of its material structure.   It's not
something I'd know how to build.
A: Protons and neutrons are in orbitals within the nucleus which  have angular momentum , so the statement of "stationary charges" is true only to first order. 
The magnetic fields in laboratory experiments are not strong enough to induce a proton or a neutron to exit the nucleus. 
In astronomical observations, neutron stars and magnetars are studied and there the magnetic fields are strong enough to change the shape of an atom and affect the nucleus of atoms.

For nuclei in the iron region of the nuclear chart it is found that fields in the order of magnitude of $10^{17}G$ significantly affect bulk properties like masses and radii. 

It is possible that if stronger astrophysical fields exist, the nucleus may break apart due to the magnetic field. This is studied in astrophysics  as the "Coulomb breakup" of the nucleus, for example here. 
A: One way this could happen is if you have a highly relativistic (very fast) nucleus travelling in a large magnetic field, such that in the nucleus' rest frame the magnetic field is partially transformed into an electric field. For a very rough calculation, I'll say that the force required to remove a proton from a nucleus is  $F_{strip} = 25 \textrm{ kN}$ (based on the plot from the wikipedia page for the nuclear force, because my nuclear physics knowledge is nearly non-existent). Assuming the nucleus is travelling perpendicular to the magnetic field, the field magnitude is $E = \gamma v B$ in the nuclear rest frame, and we can just multiply by the proton charge to get the force, then substitute a value of $B$ representing the highest known magnetic fields in the universe, magnetars at around $10^{11} \textrm{ T}$:
\begin{equation}
F_{strip} = \frac{qvB}{\sqrt{1-v^2/c^2}}, \quad v^2 = \frac{\left(\frac{F_{strip}}{qB}\right)^2}{1+\left(\frac{F_{strip}}{qBc}\right)^2}
\end{equation}
Doing these substitutions gives $v/c = 0.999999981568$, or a Lorentz factor of $\gamma \approx 5200$ which is surprisingly feasible.
