Yes! Also, it is not the 'Planck Power' (despite what you might have read in Misner, Thorne and Wheeler, P.980). The existence of black hole horizons implies a maximum luminosity (power) limit in General Relativity. Not even gravitational waves can escape a black hole . Consider an (almost) black hole made of light (this is called a Kugelblitz ) sphere of radius \begin{equation} \notag R \geq \frac{2Gp}{c^3} \end{equation} which is filled with photons with a total mass-energy of momentum $p$ times speed of light $c$ \begin{equation} \notag E=p \ c \end{equation} that leave after a time \begin{equation} \notag t=R/c \end{equation} with average power (luminosity) \begin{equation} \notag P = \frac{E}{t}=\frac{p \ c^2}{R}=\frac{c^5}{2G} \end{equation} This is maximum power in GR , regardless of the nature of the system. You might be tempted to call this half a 'Planck Power' but there is no $\hslash$ in this expression, it is purely classical. This is why
An easy-to-read journey spanning 100+ years of geometric algebra, quantum mechanics and relativity, right up to some of the biggest questions (and solutions) of present-day physics. Many giant shoulders stood upon.