Does the electromagnetic field have a "rest mass" that is conserved? In an answer to this Physics SE question, @ChiralAnomaly demonstrated that, indeed, there is a minimum field energy density observable at any point in an EM field.  With a bit more calculation, it's easy to show that if we define the usual field invariants as $k_1$ and $k_2$ where $$k_1 = (|E_p|^2 -|B_p|^2)$$ and $$k_2 = |E_p| |B_p|$$ and $E_p$ and $B_p$ are respectively the values of $\vec {E}$ and $\vec{B}$ in any inertial frame in which $\vec {E}$ and $\vec{B}$ are parallel, then the minimum field energy $H_0$ is: $$H_0^2 = (k_1)^2 + 4(k_2)^2.$$  Because $H_0$ is composed of field invariants, it is also a field invariant. Just for fun, I'll call $\frac{H_0}{c^2}$ the "rest mass density" of the field.
Here's my question: Is the integral of $\frac{H_0}{c^2}$ over all space conserved?
$H_0$ is invariant with respect to Lorentz transformations, but Lorentz invariance of a quantity that has the dimensions of energy (eg, a Lagrangian) does not imply that the quantity is conserved. Such a quantity might be conserved, but it might not.  I think $\frac{M_0}{c^2}$ is an example of such a quantity that is conserved ($M_0$ is rest mass density of, e.g., a fluid).  
What I would like to know is whether or not there is something resembling a continuity equation for $H_0$: if $H_0$ decreases in one place, does it increase in another place, more or less the way $\frac{M_0}{c^2}$ does?  
Edit #2: I have tried taking the 4-gradient of H, but run into terms like $(E \cdot \nabla) E$ and $(E \cdot \nabla )B$, and don't know what to do with them. 
I'm hoping to find an equation that shows "where changes in $H_0$ go to or come from", more or less the way that this equation shows where changes in the field energy density "go to or come from":
$$
\vec E\cdot\vec j=\nabla{(\epsilon_o c^2\vec B\times\vec E)}-\frac
{\partial}{\partial t}(\frac{\epsilon_o c^2}{2}\,\vec B\cdot\vec B+
\frac{\epsilon_o}{2}\,\vec E\cdot\vec E). $$
 A: 
Is the integral of $\frac{H_0}{c^2}$ over all space conserved?

No. As a counterexample, consider an oscillating LC circuit. (If you like, you can have the inductor and the capacitor overlap in space so that their fields are parallel.) At a time when the capacitor has zero field, we have $k_1<0$ and $k_2=0$ everywhere, so the integral is negative. At a time when the inductor has zero field, $k_1>0$ and $k_2=0$ everywhere, so the integral is positive.
In general, we can have systems that include EM fields and act like they have a rest mass, but in those systems the fields' contribution to the rest mass isn't given by the integral of $H_0$. For example, a box full of photons (a photon gas) does act like it has a rest mass equal to the energy of the photons, but $H_0$ will average to zero for this configuration. In relativistic terms, perfect fluids do have a comoving frame, but in the comoving frame, their stress-energy tensor doesn't usually look like the stress-energy of matter at rest (a pure energy density with no pressure).
A: Here's another counterexample.
$$\vec{E}=\hat{x}E_0 \sin \omega t \cos k z$$
$$\vec{B}=-\hat{y}E_0 \cos \omega t \sin k z$$
It is just a superposition of two transverse EM waves moving in opposite directions on the z-axis. I'm using $c=1$ units.
Here the scalar $k_2=\vec{E}\cdot\vec{B}=0$. So your scalar field that you are hoping has a conserved spatial integral is 
$$H_0=|\vec{E}|^2-|\vec{B}|^2=E_0^2 \left(\sin^2\omega t\cos^2 kz-\cos^2\omega t\sin^2 kz\right)$$
or actually you may prefer the absolute value of this, but it won't matter.
Integrate over a wavelength in the z direction, and any fixed length in the x and y directions. Whatever is happening in this volume $V$ that you are integrating over is happening identically in any other adjacent volume involving a wavelength so you can't claim the $H_0$ is flowing somewhere else.
After integration you get
$$\int H_0 dV=\frac{1}{2}E_0^2 V\left(\sin^2\omega t-\cos^2\omega t\right)$$
This oscillates sinusoidally in time so it is definitely not conserved.
If you instead considered the energy density, which is not a scalar, there would be a plus sign, so this integral would be a constant which reflects the fact energy is conserved.
