= Solution
Put $\epsilon=|E|>0$. On the energy surface,
$$
\frac{p^2}{2m}=\frac1{|q|}-\epsilon\geq0,
$$
so
$$
\boxed{|q|\leq q_{\max}=\frac1\epsilon.}
$$
Equality occurs at a turning point, so this is the smallest possible bound.
Using the natural collision continuation through the singular point, a full orbit runs between both turning points. Hence
$$
I=\frac1{2\pi}\oint p\,dq
=\frac2\pi\int_0^{1/\epsilon}
\sqrt{2m\left(\frac1q-\epsilon\right)}\,dq.
$$
Set $q=x^2/\epsilon$. Then
$$
\int_0^{1/\epsilon}\sqrt{2m\left(\frac1q-\epsilon\right)}\,dq
=2\sqrt{\frac{2m}{\epsilon}}
\int_0^1\sqrt{1-x^2}\,dx
=\frac\pi2\sqrt{\frac{2m}{\epsilon}}.
$$
Therefore
$$
\boxed{I=\sqrt{\frac{2m}{|E|}}.}
$$
If instead one identifies the singularity as a reflecting endpoint and regards one half-line as the orbit, the action is half this value; the following scaling is identical.
Adiabatic invariance keeps $m/|E|$ constant. Thus doubling $m$ doubles $|E|$:
$$
\boxed{|E|_{\rm final}=2|E|_{\rm initial},
\qquad E_{\rm final}=2E_{\rm initial}.}
$$
The energy becomes twice as negative.
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