I know that a cut boson propagator is replaced with the mass shell delta function. But what happens when you cut a fermion propagator? Do you just replace the denominator with a mass shell delta function, and do nothing to the numerator? Why? If so, it's a bit peculiar because the numerator actually reduce the degree of singularity and might alter the behavior of the poles in the complex plane.
It's correct that you only replace the denominators $1/(p^2-m^2+i\epsilon)$ by $-2\pi i \delta(p^2-m^2)$ in the propagators to compute the discontinuities. The fermionic propagators must first be rewritten so that they contain the denominator I just mentioned. You're right that the numerator isn't affected in the Cutkosky rules.
In some formal sense, you could also write the discontinuity of the fermionic propagators as $2\pi i \delta(p_\mu \gamma^\mu -m)$ but whenever there is a confusion, it just means the same thing indicated by the procedure in the previous paragraph.
Indeed, the fermionic propagators only have a "simple pole" near the mass shell: that's related to the fact that the propagators are the "inverse differential operators" and the equations for fermions are first-order rather than second-order as they are for bosons. This different "degree of divergence" near the mass shell is reflected by the Cutkosky rules.
However, just to be sure and avoid a potential misunderstanding that may be implicitly included in your question, the numerators don't really "annihilate" the delta-function. The function $x\,\delta(x)$ vanishes because $x=0$ at the only point where the delta-function has a support: $f(x)\delta(x) = f(0)\delta(x)$. However, $x\,\delta(x)$ is not what we see in the fermionic cutting rules.
Instead of $x$, the numerator is is $p_\mu \gamma^\mu - m$ or something like that. This numerator "formally vanishes" when it acts on a solution of the Dirac equation. However, in the cutting rules, you compute the whole matrix and not just its action on a specific spinor. And the matrix doesn't vanish on the mass shell: only two of its four eigenvalues are set to zero in the combination. So you still get a nonzero result.