From what I'm guessing a keyhole contour is one that looks like this
and because it can be shown that the contribution from $C_R$ and $C_\epsilon$ vanishes as $R\to \infty$, a Hankel contour looks like this.
If so, when I'm normally using a keyhole contour to solve improper integrals with bounds at infinity can I start it with something like this? $$\int_{\mathscr{H}}\frac{\ln z}{z^a(1+z)}\ dz=\int_{0}^{+\infty}\frac{\ln x}{x^a(1+x)}\ dx-\int_{0}^{+\infty}\frac{\ln x +2\pi i}{x^ae^{2\pi ia}(1+x)}\ dx$$ Where $\mathscr{H}$ is a Hankel contour surrounding $[0,\infty)$ in the clockwise sense.
Your observation is correct in principle. However, there's a good reason to start from the keyhole contour rather than the Hankel contour - the Hankel contour is not closed. If you wanted to actually evaluate any of the integrals you've written down, you'd have to evaluate the contour integral itself, and for this you'd want to apply tools like the residue theorem. You can do this with a keyhole contour but not directly with a Hankel contour.