I am not going to disagree with Lubos, because his answer is mostly correct, but the quantities in the path integral can be also be interpreted as operators on the Hilbert space of states, if you like. They are classical quantities on each individual trajectory of the path integral (for bosonic fields), but they become operators after you integrate, when they are sitting inside the integral sign.
The state space of a path integral is defined by superpositions on the boundary conditions. If yuo multiply by some insertion A(x,t) in the integral, you are mixing up the superpositions when the integral hits that time by multiplying by a different quantity on each path. The mixing up is a linear operator on the boundary conditions, and it is exactly the linear operator A(x,t) in Heisenberg picture quantum mechanics.
For fermionic fields, they are always "operators" in some sense, because they are always anticommuting. But their anticommutation relation is independent of the dynamics in the path integral expansion, and reduces to classical Grassman variables. Multiplying by the Grassman field inside the path integral has the same effect on Grassman coherent states as the corresponding Heisenberg picture operator.
To give an example, consider the operator X(t). This is an operator in quantum mechanics, and it obeys the canonical commutation relation:
$$[X(t),P(t)] = i$$
Inside the path integral, X(t) is just a number on each trajectory, and P(t) is also a (divergent) number in the Lagrangian path integral. A quantum state $\psi(x)$ at time t_0 is described by the superposition over initial states
$$\psi(z,t) = \int dy \psi(y) \int_{x(t_0)=y}^{x(t)=z} e^{i\int_{t_0}^t {1\over 2} \dot{x}^2 - V(x)} Dx$$
Multiplying by X(t_0) has the effect of rearranging the initial condition wavefunction into
$$\int dy X(y)\psi(y) \int_{x(t_0)=y} r^{i\int_{t_0}^t {1\over 2} \dot{x}^2 - V(x) } Dx$$
And this is exactly the same as multiplying by the operator X. To recover the commutation relation, notice that
$$X(t)V(t)$$
is ambiguous, because it depends on the time order which you use to resolve the product:
$$X(t)V(t+\epsilon) = \hat{V}(t)\hat{X}(t)$$
where the right hand side is the operator product as matrix elements, and this is justified because you multiply the initial conditions by X(t) first, then later you multiply them by P(t),
$$X(t)V(t-\epsilon) = \hat{X}(t) \hat{V}(t)$$
Where again the right hand side is an operator product, and the left side are the matrix elements of this product. The difference between the two is nonzero, because the paths are not differentiable, $\Delta X^2$ is proportional to $\epsilon$, not $\Delta X$. So that:
$$ X(t+\epsilon)V(t) - X(t)V(t) = {(X(t+\epsilon) - X(t))^2\over \epsilon} = i $$
Where the "i" is a 1 in Euclidean space, the velocity is a forward (Ito) difference, and so is always slightly ahead in time, and the last equality is a weak equality, valid only in the sense that the average over a small interval of the left and right side are equal (or equal in the sense of distributions), and valid only in the limit of approaching real time from Euclidean time, so that the oscillatory integrals are controlled.
The lack of differentiability is the same as in stochastic processes derived from Brownian motion, the square of the deviation is proportional to $\epsilon$, not like for differentiable functions, where the deviation itself is proportional to $\epsilon$.
This way of looking at things, where the quantities inside the path integral are operators, was used by Schwinger, and he liked it because it incorporated fermions naturally. Today we use Grassman integrals for the same purpose. The non-commutativity of the products is always there, however, and must be taken into account.
