I would like to try and answer my own question, which I was able to do taking hint from Connor's answer. The essential point is that the differential equation $f''(x) = 0$ has a qualitatively distinct solutions from $f''(x) = -\lambda^2 f(x)$ for $\lambda > 0$. It is, in fact, a special case that we need to treat separately.
We will look for zero modes of the Lagrangian in the OP. We assume a globally hyperbolic spacetime with Cauchy surface $S$ to serve as a setting for an initial value problem. We also assume a timelike Killing vector $K$ which serves as our time coordinate, with respect to which the spatial coordinates are orthogonal. This implies the metric satisfies
$$\partial_0 g_{\mu\nu} = 0, \quad g_{0i} = 0.$$
where latin indices $i, j$ imply spatial coordinates. The Euler-Lagrange equation for $\mathcal{L}$ gives
$$\square\phi \equiv g^{\mu\nu} \nabla_\mu \nabla_\nu \phi = 0$$
where $\square$ is the D'Alembertian. Following the procedure from Sean Carroll's Spacetime and Geometry (Sec. 9.4, pg 400), we can incorporate our conditions on $g_{\mu\nu}$ above to write
$$\partial_0^2 \phi = L_{\mathbf{x}}(\phi)$$
where $L_{\mathbf{x}}$ is a 2nd-order differential operator with respect to spatial coordinates. Precisely because of the massless setting, (and no coupling with the curvature), $L_{\mathbf{x}} (\phi) = 0$ if $\phi$ has no spatial dependence. As we often do, we will seek a separable solution $\phi(t, \mathbf{x}) = T(t) X(\mathbf{x})$, where in particular the time portion satisfies
$$T''(t) = -\omega^2 T.$$
Setting $\omega = 0$ defines what we mean by searching for a zero mode, in this context. The spatial part of the zero mode satisfies $L_{\mathbf{x}} (X) = 0$, which can be satisfied by assuming $X = \text{const}$. Solving the simple equation for the time part $T$, we find that the zero mode $\phi_{\rm ZM}$ satisfies
$$\phi_{\rm ZM} (t) = \dot{\phi}_0 t + \phi_0 $$
for initial conditions $\dot{\phi}_0$ and $\phi_0$.
The major point, as Connor pointed out, is that $\dot{\phi}_{\rm ZM}$ is not zero. Hence, neither is the conjugate momentum $\pi_{\rm ZM}$, though both $\phi_{\rm ZM}$ and $\pi_{\rm ZM}$ lack spatial dependence. Upon quantization, the only natural choice for equal-time commutation relations is
$$[\phi_{\rm ZM}(t), \pi_{\rm ZM}(t)] = i.$$
Note the lack of spatial dirac delta $\delta(\mathbf{x} - \mathbf{x'})$, which makes sense because if the classical fields have no spatial dependence, neither should the quantized versions. Therefore, this zero mode provides a representation of a single harmonic oscillator.