I once watched an online seminar by Robert Spekkens, who said something along the lines that no-go theorems are interesting, because they put constraints on what an epistemic interpretation of quantum mechanics can look like. Any no-go theorem makes a certain set of assumptions, and if the theorem is correct, we know that we must avoid at least one of those assumptions if we're to make a successful theory.
The Pusey-Barrett-Rudolph paper spells out some of its assumptions. (They do it most explicitly in the concluding paragraphs.) There may well be additional unmentioned assumptions (e.g. causality), but the ones they specifically mention are:
There is an objective physical state $\lambda$ for any quantum system
There exists some $\lambda'$ that can be shared between some pair of distinct quantum states $\psi_1$ and $\psi_2$. That is, $p(\lambda=\lambda' | \psi=\psi_1)$ and $p(\lambda=\lambda' | \psi=\psi_2)$ are both non-zero. (This is what it means for an interpretation to be epistemic, according to Spekkens' definition.)
The outcomes of measurements depend only on $\lambda$ and the settings of the measurement apparatus (though there can be stochasticity as well)
Spatially separated systems prepared independently have separate and independent $\lambda$'s.
It is from these that they derive a contradiction. Any theory that fails to make all of these same assumptions is unaffected by their result.
I'm fairly sure that under the standard formulation of Bohmian mechanics violates the second one, i.e. Bohmian mechanics is not an epistemic theory in the sense of Spekkens. This is because in Bohmian mechanics the physical state $\lambda$ consists of both the particle's real position and the "quantum potential", and the latter is in a one-to-one relationship with the quantum state.
If I understand correctly, you're suggesting that you could instead think of the physical state, $\lambda$, as consisting only of the positions and velocities of the particles, with the non-local potential being considered part of the equations of motion instead. But in this case, the third assumption above is violated, because you still need to know the potential, in addition to $\lambda$, in order to predict the outcomes of measurements. Since this formation would violate the assumptions of Pusey, Barrett and Rudolph's argument, their result would not apply to it.
In your updated question, you clarify that your suggestion is to fix the wavefunction to a particular value. In that case it's surely true that Bohmian mechanics reduces to particles moving according to deterministic but non-local equations of motion. But then you have only a partial model of quantum mechanics, because you can no longer say anything about what happens if you change the wavefunction. My strong suspicion is that if you take this approach, you will end up with an epistemic model, but it will be a model of only a restricted subset of quantum mechanics, and this will result in Pusey et al.'s result not being applicable.
Clearly no one can say that "Bohmian Mechanics is incompatible with relativity". They can say that "no known reconciliation is known to exist" but even that is questionable. Sheldon Goldstein has been working on ways to make Bohmian mechanics relativistic for awhile with varying success https://arxiv.org/abs/1307.1714
As Luke said in the comment this https://arxiv.org/abs/quant-ph/0105040 is another paper where Goldstein tries to make a backwards causal theory.
There is no widely accepted (by physics community as a whole) reconciliation as of yet between Bohmian mechanics and relativity, but you're not alone in thinking, as Goldstein does, that the answer could lie somewhere with backwards causation. This could also fix up the creation and annihilation problems for Bohmian mechanics since the electron positron creation/annihilation could be viewed as simply a particle that swapped temporal directions as opposed to "disappearing" or "appearing" out of nowhere.
So there is potentially plenty of meat on this bone and hardly anyone working on it. You should try to flesh it out yourself if you are equipped with the knowledge. Might be a real revolutionary possibility here that is being left unattended.
Best Answer
It does not "disprove" hidden variables, but rather says that, under a suitable formulation and assumptions about what form the hidden variable theory takes, that the hidden variables must be at least as complex as the quantum Hilbert space. That is to say, there is a(n almost) distinct class of hidden-variable assignments for each and every of the uncountably many, $\beth_1$, possible vectors in the Hilbert space or, if you like, that the hidden variables can be described or labeled in such a way that at least one of those variables must be equivalent to the quantum vector itself.
The idea behind the PBR theorem is this. Consider a coin in classical probability theory - not one we're going to flip, but one we're going to have some other device give us in either the heads or tails conditions, randomly, according to some settings. If that device is set up to choose heads with some probability $p$, then a person ignorant of exactly what the coin has been flipped to, but knows the setting on the device, must say that it is heads only with probability $p$. Moreover, by setting $p$ suitably on the device, we can convey to someone else a "heads" with any real-number probability $p$ even though the coin itself has only two states.
Then, under suitable combining assumptions for how multiple coins' states combine, if we have a preparation system for a quantum "coin", the theorem tells us we cannot treat the whole continuum of quantum probabilities about it as simply being due to that it has a few states and the preparation system generates those probabilities by mixing them together. Instead, it is like the coin really has not only infinitely many states, but so many there is at least one for each analogue of the real number $p$ in the classical coin case (for a qubit, this means a point on the Bloch sphere) or to say it another way, it is as though the coin carried the whole probability setting that was on the preparer. Or else, either the combining assumptions are wrong (though I've also heard that isn't "strong enough" to defeat the theorem), or, in a result reminiscent of Bell's Theorem but for the preparer, multiple preparers must somehow "spookily" correlate their preparations non-locally.