Why can't branch cut pass through poles? In the wiki article, Example (IV) – branch cuts. Why can't we can't we choose the contour so that the branch cut is on the negative x axis. If we choose this, the two residual is out of the contour, that seems good, but...
 A: Of course you can choose the branch cut on the negative axis.  It just takes a little more thinking.  Let's work that example using a branch cut on the negative axis and see what happens.
To save people the trouble, the problem is to compute
$$\int_0^{\infty} dx \frac{\sqrt{x}}{x^2+6 x+8} $$
To compute this integral using residues, we use a keyhole contour, but about the negative real axis.  To this effect, we consider the contour integral
$$\oint_{C_-} dz  \frac{\sqrt{x}}{x^2-6 x+8} $$
where $C_-$ is the keyhole about the negative axis.  Note that I had to change the denominator.  You will see why as we write out the contour integral, which is 
$$e^{i \pi} \int_R^{\epsilon} dx \frac{e^{i \pi/2} \sqrt{x}}{x^2+6 x+8} + i \epsilon \int_{\pi}^{-\pi} d\phi \, e^{i \phi} \frac{\sqrt{\epsilon} e^{i \phi/2}}{\epsilon^2 e^{i 2 \phi}-6 \epsilon e^{i \phi} + 8} \\ + e^{-i \pi} \int_R^{\epsilon} dx \frac{e^{-i \pi/2} \sqrt{x}}{x^2+6 x+8} + i R \int_{-\pi}^{\pi} d\theta\, e^{i \theta} \frac{\sqrt{R} e^{i \theta/2}}{R^2 e^{i 2 \theta}-6 R e^{i \theta} + 8}$$
Note that the branch cut along the negative axis forces the upper branch of the keyhole to have argument $+\pi$ and the lower to have argument $-\pi$.  We had to choose the denominator of the contour integral as we did to reflect the fact that were are integrating along the negative axis.  (Or above and below the branch cut.)
Now take the limits as $R \to \infty$ and $\epsilon \to 0$; the fourth and second integrals vanish, respectively.  Thus, the contour integral is equal to
$$i 2 \int_0^{\infty} dx \frac{\sqrt{x}}{x^2+6 x+8} $$
I hope you appreciate that, to use the negative real axis as a branch cut, we had to alter the integrand of the contour integral.  It wasn't rocket science, but it sure is a bit more difficult than using the positive branch!
Now we may invoke the residue theorem, which states that the contour integral is equal to $i 2 \pi$ times the sum of the residues about the poles of the integrand.  Now, keep in mind that the poles are now on the positive axis, as $z=2$ and $z=4$.  Thus, the contour integral is also equal to
$$i 2 \pi \left [\frac{\sqrt{2}}{2 (2-3)} + \frac{\sqrt{4}}{2 (4-3)} \right ] = i \pi (2-\sqrt{2})$$
Equating the two results, we get that
$$i 2 \int_0^{\infty} dx \frac{\sqrt{x}}{x^2+6 x+8} = i \pi (2-\sqrt{2})$$
and you'll find that we get the result derived in the linked example.
I noticed the title to your question is "Why can't branch cuts pass through poles?"  This is not what is happening.  I hope the above answers your actual question.
