Since Einsteins GR tells us all the frames of reference are equal, is there anything invalid about treating the Earth as unmoving and the universe itself rotating?
Other than the fact that the mathematical model is much more complex, is there anything that is wrong with it?
It's an oversimplification to say that GR treats all frames if reference as equal. In particular if we take any frame (strictly speaking any coordinate system) we can calculate the proper acceleration of an observer at rest in the frame and the result will be zero or non-zero depending on the frame. If the proper acceleration is zero then the frame we have chosen is locally equivalent to an inertial frame, while if the proper acceleration is non-zero the frame is locally equivalent to a non-inertial frame.
So for example if we take the frame of an observer at rest on the surface of the Earth this frame is locally non-inertial. That means freely moving objects won't move in straight lines i.e. if you throw a stone it will move in a curve (approximately a parabola) and a pendulum will rotate its plane of swing by $2\pi$ every $24$ hours. From a Newtonian perspective there will be fictitious forces acting.
Even in Newtonian mechanics there's nothing wrong with using non-inertial frames. They are just (as you say) more complicated to do calculations in. Likewise in GR there's absolutely no problem in choosing a frame corotating with the Earth, but it will make any attempts to do calculations more complicated than they need be. For example in this frame the flat spacetime metric:
$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$
$$ ds^2~=~\Big(1~-~\frac{r^2\Omega^2}{c^2}\Big)dt^2~-~r^2d\theta^2~-~2r^2\frac{\Omega}{c} d\theta dt~-~dr^2 $$
Yes! Every coordinate system in GR is "treated as equal". But the quantities you measure, by means of "clocks and rods" (of course with much more complicated devices), are independent on your choice of coordinates. Like the metric itself; indeed the metric $g_{\mu\nu}$ is NOT measurable, since is an object which depends on coordinates (as in your case the coordinates by which the earth is not rotating). But the $\textit{space-time interval}$ $$ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}$$ is instead measurable since it is a scalar. Now, this is an over-semplification, since what I just wrote holds only locally, but there are several spacetimes in which "local" means "global" since I can cover the entire space-time manifold with just one chart (system of coordinates).
