Because $|\phi|^2=\rho$ is the boson density, and you expect to have finite boson density in the Mott phase (like one boson per site), so you can not just let the boson field $\phi$ goes to zero.
In fact, there are two ways to disorder a superfluid and restore the U(1) symmetry, either by the amplitude fluctuation or by the phase fluctuation. The $r$-reversal process you described is to introduce the amplitude fluctuation, which usually corresponds to a finite temperature transition to the thermal boson gas. In this case, $\phi$ describes the condensed part of the boson, and decreasing the amplitude $|\phi|$ corresponds to depleting the boson from the condensate to the thermal gas. However the superfluid-Mott transition is a zero temperature quantum phase transition. At zero temperature, there is no thermal boson to absorb the lost of $|\phi|^2$, so if you send $|\phi|$ to zero, the boson will be physically lost from the system, and the resulting U(1) symmetric state is actually the vacuum state. We know this should not be the case for the superfluid-Mott transition, as the boson density $|\phi|^2$ is fixed at a non-zero level in both phases, so the transition is not driven by the amplitude fluctuation, but actually driven by the phase fluctuation without $|\phi|$ going to zero.
The correct field theory for the superfluid-Mott transition is the XY model. In this theory, one must first separate the boson field into the amplitude and the phase modes $\phi=\sqrt{\rho}\mathrm{e}^{\mathrm{i}\theta}$. Substitute into the $\phi^4$-type of theory and integrate out the small amplitude fluctuation, we obtain the low-energy theory for $\theta$
$$\mathcal{L}=\frac{\eta}{2}(\partial_\mu\theta)^2,$$
where $\eta$ is the superfluid stiffness. Because $\theta$ is a compact variable (meaning that it has the periodicity $\theta\sim\theta+2\pi$), it is more convenient to consider its dual field theory in terms of $\psi$, such that $\epsilon^{\mu\nu}\partial_\nu\psi$ is the conjugate variable of $\partial_\mu\theta$, with $v\equiv\mathrm{e}^{\mathrm{i}\psi}$ being the annihilation operator of the vortex in the superfluid,
$$\mathcal{L}=\frac{1}{2\eta}(\partial_\mu\psi)^2+\frac{g}{2}(v+v^\dagger)=\frac{1}{2\eta}(\partial_\mu\psi)^2+g\cos\psi.$$
In the large $\eta$ limit, $(\partial_\mu\psi)^2$ costs little energy, so $\psi$ field fluctuates strongly (and $g\cos\psi$ is averaged to zero and hence irrelevant), then the dual variable $\theta$ will not fluctuate, which corresponds to the U(1) breaking superfluid phase. In the small $\eta$ limit, the fluctuation of $\psi$ is suppressed, and the $g\cos\psi$ term becomes relevant, which leads to the vortex proliferation (creation and annihilation of vortices). So the long-range order of $\theta$ is destroyed, and the U(1) symmetry is restored, which corresponds to the Mott insulating phase.
