It all depends on which system you are describing.
First-quantized system
For example, it is possible to quantize the following (first quantized) action for the relativistic point particle:
$$ S[X] = -m c \int d{\lambda} \sqrt{g_{\mu \nu} \dot{X}^{\mu} \dot{X}^{\nu}}, $$
where $\dot{X}$ stands for $dX/d\lambda$ and $\lambda$ is some arbitrary unphysical parameter used to label the points of the worldline with real numbers.
When you quantize this system, your kinematical (unconstrained) Hilbert space $\mathcal{K}$ is a space of rapidly decreasing spacetime functions:
$$ \Psi(t,\mathbf{r}) \in \mathcal{K}. $$
Just like in the nonrelativistic case, the uncertainty relations read
$$ \Delta{x} \Delta{p} \le \frac{\hbar}{2}, $$
but there's also another relation
$$ \Delta{t} \Delta{E} \le \frac{c \hbar}{2}$$
which we encountered earlier as the time-energy uncertainty relation, only in this description it is a first-class citizen.
However, the picture is obscured by the existence of constraints. The physical Hilbert space $\mathcal{H}$ is given by those elements of $\mathcal{K}$ which are solutions of the Klein-Gordon equation. (To be precise, by elements of $\mathcal{K}^{*}$ which vanish when evaluated on solutions of Klein-Gordon equation).
Second-quantized system
Or you can jump right to Quantum Field Theory and inspect the Klein-Gordon lagrangian as a Lagrangian for the quantum field. You will then have wavefunctionals
$$ \Psi[\phi(\mathbf{r})] $$
as states, and the uncertainty relation reads
$$ \Delta \phi \Delta \pi \le \frac{h}{2} $$
with $\pi(\mathbf{r})$ the canonical momenta for $\phi(\mathbf{r})$.
This description is considered more fundamental. The position-momenta and time-energy uncertainty relations for elementary particles follow from this if one considers the fluctuations of the vacuum state of the field (elementary particle states).
