I'm going to lean heavily on the "or any such expression" part of your question. Because there's an interesting way of formulating QFT along these lines, but I'm totally leaving behind your idea that $x$, the position in space, might return as an operator. I'm not aware of any such formulation in QFT.
Quantum field theory on a lattice is often done with a formalism like this. See for example the first formulation of the Jordan, Lee, Preskill algorithm, an algorithm for simulating scalar field theory with a quantum computer. At least scalar quantum field theory can be formulated like this:
The state of a system is a given by a "wave-functional" (something that takes in a function of spacetime and returns a complex number)
$$
|\Phi\rangle=F[\phi(x)]
$$
To reiterate, I can put in a proposed form for the classical field like $\phi(x)=e^{-|x|^2}$ and my wavefunctional will map this to a complex number whose magnitude kind of denotes the probability that if I somehow measured the field throughout all of space I would get exactly that result.
Now $\phi(x')$ is an operator the same way $x$ is an operator in quantum mechanics done with wavefunctions... it literally just gets multiplied with a functional:
$$
\hat{\phi}(x')|\Phi\rangle=\phi(x')F[\phi(x)]
$$
Just to give you an idea of how this works, let's calculate an expectation value of the field at a particular point in space $x'$... We're going to need to integrate over the Hilbert space and its going to be a functional integral.
$$
\langle\phi(x')\rangle=\langle\Phi|\hat{\phi}(x')|\Phi\rangle
$$
$$
=\int\cdots\int\cdots\int
\\F^*(\phi(x_1),\cdots,\phi(x'),\cdots,\phi(x_n))\\\phi(x')\\F(\phi(x_1),\cdots,\phi(x'),\cdots,\phi(x_n))\\d\phi(x_1)\cdots d\phi(x')\cdots d\phi(x_n)
$$
Note those five lines are all one line spaced out to make different components more visually apparent... please don't get confused by that. And I've tried to make it clear that the wavefunctional $F[\phi(x)]$ can be thought of as a function independently of the field value at every point in space. In fact if you're on a finite lattice, then it is just a function of a finite number of variables. So now finally what about canonical commutation relations? Well the conjugate momentum to $\hat{\phi}(x)$ is given by:
$$
\hat\pi(x)=-i\hbar\frac{d}{d\phi(x)}
$$
Fields are the operators in quantum field theory, not position. The conjugate momentum operator is a derivative with respect to the field value, not $d/dx$.
Say you want to know something like the groundstate of a quantum field theory... you need to find the eigen-functionals of the hamiltonian, say for $\phi^4$ interacting QFT ($i=c=\hbar=-1=1$):
$$
\hat{H}=\int_{dx}\hat{\pi}^2(x)+\hat{\phi}^2(x)+\lambda\hat{\phi}^4(x)
$$
Why isn't this formalism used more often? Because it only makes any sense on a lattice with finitely many space points. Otherwise the functional is totally riddled with infinities and zeros (even in noninteracting QFT). It returns zero for every field input, and it returns infinity for everything you might try to calculate.
