Nothing would change in terms of $\mathbf{E}$ and $\mathbf{B}$ fields, but there is a reason why the generally the Lorenz gauge is preferred: it is Lorentz (with a t) invariant! In other words, if you take a $\phi$ and an $\mathbf{A}$ that satisfy the Lorenz condition, the $\phi'$ and $\mathbf{A}'$ that another inertial observer sees will also satisfy it. This is not the case for the other condition that you have provided with the minus sign.
If you want to take a deeper look, the reason why the condition is written like that is that it states nothing more than the fact that the four-potential $A^{\mu} = (\phi, \mathbf{A})$ has vanishing divergence: $$ \partial_{\mu} A^{\mu} = 0. $$
Writing this out more explicitly gives us the Lorenz gauge:
$$ \partial_{\mu} A^{\mu} = \partial_{0} A^0 + \partial_{k} A^{k} = \frac{1}{c} \frac{\partial \phi}{\partial t} + \nabla \cdot \mathbf{A} = 0.$$
Generally, you can invent any gauge you want (as long as there are potentials that can satisfy it), but the question is whether you can find a use case for it. The Lorenz gauge is useful in more abstract situations where we would like a Lorentz invariant condition, but in the context of electrostatics, the Coulomb gauge ($\nabla \cdot \mathbf{A} = 0$) turns out to be the most useful. In the end, the best choice of gauge depends on the problem you are trying to solve.