The field and its conjugate momentum are operators. They act on the Hilbert space of the theory, which is not the space of square-integrable functions of position, as it is in single particle quantum mechanics. Rather, the configuration space of the theory is the set of all field configurations and the wavefunction is actually a functional, which ascribes an amplitude $\Psi[\phi(x^\mu)]$ for the field to be in a configuration $\phi(x^\mu)$.
QFT is just quantum mechanics applied to an infinite collection of degrees of freedom labeled by positions $x$. Once you get used to the switch it comes to seem pretty simple, but the set of all field configurations makes a huge space so naturally it is very difficult to make things mathematically rigorous. For the most part physicists can live without rigourously defining the configuration space.
The dictionary from ordinary quantum mechanics to quantum field theory is
$$
\begin{array}{lcl}
\mathrm{QM} && \mathrm{QFT}\\
t,i &\to& \left(t,x,y,z\right)\equiv\left(t,{\bf x}\right)\equiv x^{\mu}\\
q_{i}\left(t\right) &\to& \phi\left(x^{\mu}\right)\\
p_{i}\left(t\right) &\to& \pi\left(x^{\mu}\right)\\
\psi\left(t,q_{1},q_{2},\cdots,q_{N}\right) &\to& \Psi\left[\phi\left(x^{\mu}\right)\right]\\
\left[q_{i}\left(t\right),\ p_{j}\left(t\right)\right]~=~i\hbar\delta_{ij} &\to& \left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)\\
\hat{H}\left(t\right) &\to& \hat{H}\left(t\right)=\int\mathrm{d}^{3}x\ \hat{\mathcal{H}}\left(x^{\mu}\right)
\end{array}
$$
You can represent the field momenta explicitly by functional derivatives:
$$ \pi(x^\mu) ~=~ - i\hbar \frac{\delta}{\delta \phi(x^\mu)}, $$
and check that this satisfies the commutation relationship.