You are not right. There is no orthogonal basis $|x,t\rangle$ for fixed $t$ in relativistic quantum mechanics, but there is a state $|x,t\rangle=|x^{\mu}\rangle$. In quantum field theory this state is expressed by $\phi(x)|0\rangle$, but this is a one-particle state, so can be expressed in the one-particle space, spanned by the momenta eigenstates $|\vec{p}\rangle$.
$$ |x^{\mu}\rangle=(2\pi)^{-3/2}\int \frac{d^{3}\vec{p}}{\sqrt{2E_p}}e^{ip_{\mu}x^{\mu}}|\vec{p}\rangle $$
The main point is that the state $|x\rangle$ does not imply the existence of a hermitian operator $x$, so this state can exist without the existence of $x$-operator. In fact, this operator can't exist since the Hamiltonian is bounded from bellow. You can see more about it here.
If the state $|x^{\mu}\rangle$ exist, makes perfect sense to have path integrals over trajectories $x_{\mu}(\tau)$, if we gauge away the parametrization at the end.
$$ \langle x_2^{\mu}|x_1^{\mu}\rangle=\int_{x(0)=x_1}^{x(1)=x_2}\frac{\left[dx(\tau)\right]}{\mathbb{diff}}\,e^{-S[x]} $$
What you will discover from that path integral is that $\langle x_2|x_1\rangle \neq 0$ even for space-like intervals, and by <PERSON> symmetry you may deduce that the same happens for hyper-surface of simultaneity. This means that we cannot take $|x\rangle \rightarrow |\vec{x},t\rangle$ as an orthogonal basis for $t$ fixed.
This is why there is no wave function $\psi(x)$ description for the states. 66
For the momentum wave-function there will be no problem, and a wave-function $\phi(p)$ is totally acceptable. For instance, $P$ is a hermitian operator for the one-particle Hilbert space in the free theory (free means no mixing with multi-particles states).
Actually the BRST-approach is far from be just pedagogical. You can obtain the field theory description by defining a field operator $\Psi(x^{\mu})$ and impose the equation of motion:
$$ Q\Psi=0 $$
This is useful to obtain non-linear generalization (i.e. how to put interactions) for particles in the presence of supersymmetry and/or gauge symmetry. The gauge transformations of the field will be described by the generalization of the BRST-exact fields $\delta\Psi=Q\Lambda+g\left[\Psi,\Lambda\right]$ and the equation of motion will be:
$$ Q\Psi + g\Psi^2=0\,. $$
You can see more about it here. 976