Does an electric field cause the vacuum to emit photons? As I understand it electron-positron pairs pop in and out of the vacuum on a time scale $T$ inversely proportional to the electron mass.
Imagine we put a static electric field across the vacuum.
Within the time scale $T$ the electron and positron would be pulled apart a little by the electric field before they had a chance to annihilate back into the vacuum.
Would the difference between the initial size of the particle pair and the final size lead to a difference in momenta which would then be radiated as a real black body photon?
The Fulling–Davies–Unruh effect says that a body with acceleration $a$ experiences a vacuum temperature:
$$T\sim\frac{\hbar\ a}{c k_B}$$
If we have an electric field $E$ then an electron-positron pair is separated with an acceleration a given by
$$a = \frac{e E}{m_e}$$
Thus we should have a temperature $T$ given by
$$T \sim \frac{\hbar e E}{m_e c k_B}$$
For an electric field $E=10^6$ V/m we should be able to measure the electron-positron pairs radiating photons with a wavelength of $10$ cm.
PBS Space Time seems to believe that Unruh radiation from an accelerating body can be observed by an inertial observer:
https://youtu.be/7cj6oiFDEXc&t=426s
 A: To the extent that there is production of particles by a constant electric field form the vacuum, you're just talking about the Schwinger effect, even if you think you aren't.

As I understand it electron-positron pairs pop in and out of the vacuum on a time scale $T$ inversely proportional to the electron mass.

This is just a pop-sci interpretation of quantum field theory that bears little to no resemblance to what actually happens in the formalism. It is a common misinterpretation of Feynman diagrams like vacuum bubbles as depicting actual "processes" rather than computations in a perturbation series (see also e.g. this answer of mine and this answer of mine for further discussions on this point). Thus your attempt to apply the Unruh effect to virtual electron-positron pairs doesn't make any sense - even apart from the issue that non-accelerating observers don't see the particles created from the Unruh effect.
The real way to compute production of anything from the vacuum must works as follows: We start with a vacuum $\lvert 0\rangle$ with no electric or other background fields applied. Then we activate the background fields for some time, and then deactivate them again to end up with a theory whose vacuum is still just $\lvert 0\rangle$. The background fields in between in principle evolve the initial vacuum $\lvert 0\rangle_\text{in}$ in time by some $U[A]$ (by which I mean the time evolution operator associated with a non-trivial EM 4-potential $A$ acting over the time between the asymptotic past and future). The question is now when
$$ \langle 0\vert U[A]\vert 0\rangle$$
is not unity, i.e. when there is a possibility that we end up not only with the vacuum $\lvert 0\rangle$ but with some state that contains actual, real particles. [Side note: The Unruh effect works not by having some $U$, but by having the final vacuum be not $\lvert 0\rangle$ but some $\lvert 0_a\rangle$ that an observer with acceleration $a$ "thinks" is the vacuum.]
This is precisely what Schwinger computes in his famous series of papers (I, II) and it leads to the well-known amplitude for the Schwinger effect. Crucially - as Schwinger himself remarks in II, no particle production

[...] will indeed result if the energy and momentum conservation laws cannot be simultaneously obeyed in the course of the pair producting interaction between the electromagnetic field and the fluctuating current in the vacuum.

That is, if you don't put enough energy into the electric field to be able to conserve energy, you won't see any electron-positron pairs nor any photons, otherwise you'll see them just at the rate predicted by the Schwinger effect. [Side note: Pay attention to the length of these papers and the carefulness of Schwinger's arguments - this is how proper arguments about quantum field theoretic effects must work.]
Note that the Schwinger effect is not actually hard-limited by the strength of the electric field - in principle you can observe pair production at very low field strengths if the "total" electric field in that region has enough energy to "pay for" pair production, it's just so vanishingly unlikely it's not feasible to hope to ever detect that. The "Schwinger limit" is simply a value where this probability gets large enough it might matter for anything, not some sort of hard cutoff.
Since your photons need to be produced by annihilation of real electron-positron pairs, they're just a "secondary stage" of the Schwinger effect, not a new effect.
