What if the two Black Holes spiraling around each other are evaporating via their Hawking radiation?
Gravitational waves are efficiently emitted by massive black holes orbiting each other - the power emitted increases with mass. Hawking radiation on the other hand is a process that increases with decreasing mass. As a result only very tiny black hole binaries would emit more power in Hawking radiation than they do in gravitational waves; at least towards the end of the inspiralling phase.
Details:
The characteristic timescale on which a black hole binary system spirals to merger is
$$\tau_{\rm GW} \simeq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{M^3}\right)\ ,$$
where $a_0$ is the separation, $M$ is the total system mass and I've assumed the binary components are of equal mass.
The characteristic evaporation timescale by Hawking radiation (again assuming each black hole is of mass $M/2$, is
$$ \tau_{\rm Evap} \simeq 640\pi \frac{G^2 M^3}{\hbar c^4}$$
For $\tau_{\rm Evap}$ to be shorter or even comparable with $\tau_{\rm GW}$, then
$$ 640\pi \frac{G^2 M^3}{\hbar c^4} \leq \frac{20c^5}{256 G^3}\left(\frac{a_0^4}{ M^3}\right),$$
which means
$$ \frac{M^6}{a_0^{4}} \leq 3.9\times 10^{-5} \frac{\hbar c^9}{G^5}$$
For black hole binaries, a reasonable value for $a_0$ would be a few times the Schwarzschild radius of the final merged black hole since most of hte power in gravitational waves is radiated in the final few orbits before merger, i.e. $a_0 \sim 10 GM/c^2$, so we can say
$$ M \leq 0.62 \sqrt{\frac{\hbar c}{G}} \sim 2\times 10^{-8}\ {\rm kg}\ ,$$
which is the Planck mass.
So the only merging black holes for which energy loss by Hawking radiation will play a significant role, close to the end of their inspiral, would have a mass of a few $\sim 10^{-8}$ kg or less. But the dynamics of such tiny black holes would not be controlled by gravity at all and they would evaporate in a fraction of a second.