The preferred formula to teach, for the Lagrangian of non-relativistic charged particle in external field, is
$$
L(\mathbf r,\mathbf v,t) = \frac{1}{2}mv^2 - q\phi(\mathbf{r},t) + q\mathbf{v}\cdot \mathbf{A}(\mathbf{r},t).
$$
This gives correct equations of motion, and can easily be recalled from the "mnemotechnic rule" that such Lagrangian should be like $T-U$ in classical mechanics, where $T$ is kinetic energy and $U$ is electric potential energy; $q\phi$ is electric potential energy, so it has to appear with a minus in $L$. The other term $q\mathbf v\cdot \mathbf{A}$ has nothing to do with energy, but it should appear there with opposite sign (compared to $q\phi$). This can be quickly recalled from the fun fact that the whole group of interaction terms in $L$ is the sum $\sum_\mu A_\mu u^\mu$ (looks like product of four-vectors, even though it isn't necessarily), in which $A_0$ is minus electric potential.
If I have the Lagrangian as
$$L = \frac{m v^2}{2} - q \vec{v} \cdot \vec{A} - q \phi$$
that gives the equation of motion as
$$m \dot{v} = q(-E + v \times B)$$
Does this equation of motion still represents the Lorentz force even with the difference in signs, or $E$ and $B$ must have the same sign?
If $\mathbf A$ is the standard vector potential (e.g. Coulomb gauge), from the proposed $L$ and the Euler-Lagrange equations, you won't get the equation $m \dot{\mathbf v} = q(-\mathbf E + \mathbf v \times\mathbf B)$, because there is no $\mathbf E,\mathbf B$ in the Lagrangian; what you will get is
$$
m\dot{\mathbf v} = q(-\nabla \phi + \partial_t \mathbf A) - q\mathbf v \times \nabla \times \mathbf A
$$
This is not the standard expression for Lorentz force, however, it can be made valid in a weird, non-standard way, where $\mathbf A$ does not denote a magnetic vector potential that obeys (the defining property) $\mathbf{B} = \nabla \times \mathbf A$ and the relation $\mathbf E = -\nabla \phi - \partial_t\mathbf A$, but it is a different thing, obeying $\mathbf B = - \nabla \times \mathbf A$ and $\mathbf E = -\nabla \phi + \partial_t\mathbf A$ instead.
Only then we can rewrite the above as
$$
m\dot{\mathbf v} = q\mathbf E + q\mathbf v\times \mathbf B.
$$
Thus if $\mathbf A$ means minus standard vector potential (which is mathematically permissible), the above Lagrangian gives the correct equations, compatible with the Lorentz force formula. But all this is in a non-standard and confusing way, and there is no upside.
Is it correct to interpret this Lagrangian as a charged particle in an electromagnetic field?
If $\mathbf A$ is minus vector potential, this Lagrangian is a non-standard, but a valid Lagrangian for a charged particle in external (fixed) EM field. It implies the correct equations of motion for the particle.
