In complex calculus, one may compute powers – and other functions – for all negative or complex values of the argument. But it would lead to confusions and ambiguities and indeed, it's not needed.
Whenever the theory is stable, it's guaranteed that $\Delta$ in these d.r. integrals ends up positive. A typical value of $\Delta$ for Feynman-parameterized integrals is an example from a 4-point diagram (imagine 2 to 2 scattering)
$$ \Delta = m^2 - x(1-x)q^2 $$
Now, $m^2$ is positive because we don't have tachyons and $x(1-x)$ is between $0$ and $1/4$. The only risk how the second term could beat $m^2$ and make $\Delta$ negative is that $q$ is hugely timelike, with length over $2m$. But it really can't happen because $q$ is the sum of two external momenta. Timelike $2m$ length is really the maximum and in most cases, $q$ will end up shorter or spacelike. The UV region we're integrating corresponds to very long spacelike $q$ (all directions in the Euclideanized spacetime are spacelike) for which $\Delta$ is high and positive because the spacelike vectors have negative signature above.
I can't provide you with a general proof (it may be easy, however) but I am pretty sure that the negative-$\Delta$ problem doesn't arise in any loop diagrams.
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