Is relativistic statistical mechanics "foundationally" relativistic?

The Maxwell-Jüttner distribution describes the distribution in a relativistic gas:

$$f(\gamma) = \frac{\gamma^2\beta}{\theta K_2(1/\theta)}e^{-\gamma/\theta}$$

with

$$\theta = \frac{kT}{mc^2}$$

The Boltzmann factor is defined in terms of energy, not velocity, and $$E = \gamma m$$, so the probability of any unbounded energy is not a problem.

Note that it is classical: it ignores quantum mechanics.

No, you cannot write that particles are allowed to surpass the speed of light but with a vanishing probability. The fundamental reason is that statistical mechanics for classical particles is constructed in phase space $$(\vec r,\vec p)$$ Therefore, in special relativity, the partition function of an ideal gas reads $${\cal Z}=z^N, \quad z=\int e^{-\beta\sqrt{p^2c^2+m^2c^4}}{d^3\vec rd^3\vec p\over h_0^3}$$ The position $$\vec r$$ is integrated over the volume and the momentum $$\vec p$$ over $$\mathbb{R}^3$$. However, $$\vec p={m\vec v\over \sqrt{1-v^2/c^2}}$$ so the speed $$v$$ remains always smaller than the speed of light.