Yes, you could consider such a dimensional reduction. The idea is to decompose $\phi(t, \mathbf{x})$ as
$$
\phi(t, r, \theta) = \sum_{k \in \mathbb{Z}} \phi_k(t, r) e^{i k \theta}.
$$
You plug this Ansatz into your action, which then becomes
$$
S = \int\!d^3x \, L[\phi] = \int\!dt r dr d\theta\, L'[\{\phi_k\}]
$$
where $L'[\{\phi_k\}]$ is a 3d Lagrangian that depends on the infinite set of modes $\{\phi_k\}$ with $k \in \mathbb{Z}$.
Finally, you will find that it's easy to perform the $\theta$ integral. After all
$$
\int\!d\theta\, e^{ik\theta} e^{in\theta} = 2\pi \delta_{k+n,0}.
$$
You will see that the kinetic term becomes something like
$$
S = \sum_k \int\!dt r dr\, \phi_{-k} (\Box_\text{2d} + k^2) \phi_k + \text{interactions}.
$$
So modes with large $k$ contribute a lot to the action and are exponentially suppressed at the level of correlation functions.
Interactions will mix modes with different $k$, in a way that's ultimately consistent with the $SO(2)$ invariance of the $3d$ theory.
What makes this dimensional reduction a bit funny is that you give up translation invariance in space. The only manifest spacetime symmetries of the final theory in the $(t, r)$ domain are time translations.
Finally, your picture of naive dimensional reduction amounts to keeping only the zero mode $\phi_0$. The full theory is described by the infinite set of modes $\{\phi_k\}$. If you want to understand this reduction quantitatively, you could truncate the theory by keeping only modes with $|k| \leq K$ for some fixed integer $K$. Using that cutoff, you would compute things like correlation functions of $\phi_0$ (which will depend on $K$). Finally, you could change $K$, which would allow you to derive a set of renormalization group equations for the couplings in your $2d$ theory. By following the renormalization group flow down to $K = 0$, you would be able to systematically compute the couplings of the naive theory in terms of those in the full theory with a cutoff. However, that computation would be the length of a small research paper, so it's not in the scope of a StackExchange answer.