lagrangian formalism - Can I really take the classical field equations at face value in QFT?

To be concrete, let's say I have a relativistic $\phi^4$ theory [with Minkowski signature $(+,-,-,-)$]

$$\tag{1} \mathcal{L} ~=~ \frac{1}{2} \left ( \partial_{\mu} \phi \partial^{\mu} \phi - m^2 \phi^2\right ) - \frac{\lambda}{4!} \phi^4.$$

The classical equation of motion for $\phi$ is:

$$\tag{2} (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 ~=~ 0.$$

I knew that canonical quantization is basically replacing all Poisson' brackets with (anti-)commutators. From that point of view, I would expect a classical field equation to remain valid as an operator equation even after quantization. Am I wrong?

If I am indeed correct, then specifically to the $\phi^4$ example, does that mean

$$\tag{3} \left \langle \left [ (\Box + m^2) \phi + \frac{\lambda}{3!} \phi^3 \right ] \mathcal{O} \right \rangle ~=~ 0$$

for any operator $\mathcal{O}$, in the full interacting theory?

And how do I reconcile this with the path integral picture?

Only the classical paths follow classical equations of motion to the letters. But to quantize a theory, every path is assigned a weight $e^{iS}$, and obviously none of these new inclusions will follow the classical equations. Then, how can the field equations still hold?