$\pi(\mathbf{x})$ doesn't generate translations of particles, it generates translations of the field, $\phi(\mathbf{x})$, at $\mathbf{x}$. At least, that's the most common way to interpret the fact that the two operators obey the equal time commutation relation:
$$[\phi(\mathbf{x}),\, \pi(\mathbf{y})] = i\hbar \delta(\mathbf{x}-\mathbf{y}).$$ Note that the delta function only covers the spatial coordinates, and the equal time commutator works in both Heisenberg and Schödinger pictures. Likewise, the fields obey the same Heisenberg picture commutation relations with the Hamiltonian that position and momentum do in ordinary QM:
$$\begin{align}
\dot{\phi}(\mathbf{x}) & = [\pi(\mathbf{x}), H] \\
\dot{\pi}(\mathbf{x}) & = [\phi(\mathbf{x}), H]
\end{align}$$
Compare the above canonical commutators with the correspondence principle relations in ordinary QM and QFT, respectively:
$$\begin{align}
\frac{\operatorname{d} \langle \mathbf{x} \rangle}{\operatorname{d} t} & = \frac{\langle \mathbf{p}\rangle}{m} \\
\frac{\partial \langle \phi(\mathbf{x}) \rangle}{\partial t} & = \langle \pi(\mathbf{x})\rangle.
\end{align}$$
The only difference between the relations is that the "mass density" that would be present in a study of classical waves in strings or membranes has been absorbed into the field definition.
As for what it means, I prefer to imagine a membrane that oscillates in a direction transverse to space-time that has a momentum density to its oscillations, though that's just a picture to help organize properties, and not really a way to ascribe "meaning."
All of the above is about what $\pi(\vec{x})$ 'does' in QFT. For a quantity of interest you can calculate in terms of a field's vector potential, look no further than quantum electro-dynamics. The fields in QED are given by the vector potential, $A_\mu(\mathbf{x})$. The gauge symmetry, and the lack of a time derivative of $A_0$ in the Hamiltonian, makes the theory behave a little strange with respect to quantization. If we pick the Weyl gauge, $A_0(\mathbf{x}) = 0$ for simplicity, then the fields are $\vec{A}(\mathbf{x})$ and the canonically conjugate momenta $\vec{\pi}(\mathbf{x})$ are defined in the usual way. The classical correspondence theorem gives is:
$$\frac{\partial \left\langle \vec{A}(\mathbf{x})\right\rangle}{\partial t} = \left\langle \vec{\pi}(\mathbf{x})\right\rangle,$$
which translates into the classical definition of the electric field as:
$$\vec{E} = - \left\langle \vec{\pi}(\mathbf{x})\right\rangle,$$ where, to reiterate, we're using the Weyl gauge where the electric potential is identically 0 everywhere.
The generator of translations for particles is, still, the momentum operator in the field. I'm a little rusty, and I don't have my textbooks handy, so I may get the operator ordering wrong, but the momentum operator for an ordinary scalar field is given by:
$$\vec{P} = -\int \operatorname{d}^3 x\, \pi(\mathbf{x}) \nabla \phi(\mathbf{x}).$$ To reiterate: $\vec{P}$ is the total ordinary space-time momentum carried by the field, and the generator of space translations of configurations of the field. $\pi(\mathbf{x})$ is the momentum density of the field in the field direction.
