When is time a degree of freedom?

What you have written even without the time derivative is wrong; remember the divergence is,

$$\nabla \cdot \vec F = \partial_x F_x + \partial_y F_y + \partial_z F_z$$

and not the derivatives acting on all of $\vec F$ as you wrote. Now, if we want to include a time derivative in the divergence, it only makes sense if $\vec F$ has a time component.

This occurs in the case of general relativity, where we may consider some vector $v^\mu$ with an index that runs over $\mu = 0, \dots, 3$ where $v^0$ is a $t$-component (or the appropriate timelike coordinate).

We could then write a divergence using the covariant derivative,

$$\nabla_\mu v^\mu = \partial_\mu v^\mu + \Gamma^{\mu}_{\mu \nu}v^\nu$$

and in the case of flat Minkowski space, this reduces to the original concept you thought of, which is simply,

$$\partial_\mu v^\mu = \partial_t v^t + \partial_x v^x + \partial_y v^y + \partial_z v^z.$$

In general, the nabla operator, in all of its form, affects only to the spatial coordinates. You must not include the time derivative in gradients, divergences, curls or laplacians. You can take them as a definition; they just don't include time derivatives.

It's true that for Special relativity and superior levels, the quantity $ct$ is considered as the 4th variable, and so the laplacian is generalized to what is called "d'Alambertian", which includes a $-\frac{\partial F}{\partial (ct)}$.

But time is independent in Classical physics. It's a "different type" of coordinate, and it must not be taken when dealing with "spatial operators".