If we integrate both sides of the Lorenz gauge condition, $\nabla \cdot \mathbf{A} = -\frac{1}{c^2}\frac{\partial \phi}{\partial t}$, over a small volume (free of charges for simplicity), we get:
$$ \int_V \nabla \cdot \mathbf{A} \, dV = -\frac{1}{c^2} \int_V \frac{\partial \phi}{\partial t} \, dV $$
Applying the Gauss divergence theorem: $$ \oint_S \mathbf{A} \cdot d\mathbf{S} = -\frac{1}{c^2}\frac{\partial}{\partial t} \int_V \phi \, dV $$
This says that the rate of decrease of $\phi$ in $V$ is proportional to the flux of $\mathbf{A}$ from the volume $V$.
The retarded potentials solution of Maxwell's equations that result from applying the Lorenz gauge condition seems to support this interpretation:
$$ \phi(t) = \frac{1}{4 \pi \epsilon_0} \int_{V_\rho} \frac{[\rho]}{r} dV_\rho \\ \mathbf{A}(t) = \frac{1}{c^2} \frac{1}{4 \pi \epsilon_0} \int_{V_\rho} \frac{[\rho \mathbf{v}]}{r} dV_\rho $$
The solutions say that two potentials are emitted and propagated from a moving charge. A strong scalar potential $\phi$, and a $\frac{1}{c^2}$ weaker, vector potential $\mathbf{A}$, that is proportional to the velocity, $\mathbf{v}$, of the moving 'emitting' charge.
If all the charges are stationary, then $\phi$ in any region is constant and so is $\int_V \phi \, dV$. If there are moving charges, the $[\rho \mathbf{v}]$ term from the weaker potential, $\mathbf{A}$, corresponds to a flux of $\phi$ that can lead to an increase or decrease of $\int_V \phi \, dV$.
Is this a correct way to interpret the Lorenz gauge condition?