If the charge is moving its position must change, so the actual time-explicit form of the charge density is:
$\rho\left(\mathbf{r},\,t\right)=q \delta^{(3)}\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)$
Next you need the $\dot{\rho}=q\frac{\partial}{\partial t}\delta^{(3)}\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)=q\frac{\partial \left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)^i}{\partial t}\,\frac{\partial}{\partial\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)^i}\delta^{(3)}\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)$
This is simply a chain rule. Note the sum over the repeated index $i$. Basically all I am saying here is that if you have a function $g=g\left(t\right)$ inside another function $f$ then
$\frac{df\left(g\left(t\right)\right)}{dt}=\dot{g}\left(t\right)\,f'\left(g\left(t\right)\right)$,
but this time instead of a single function $g$ you have three components of the $\mathbf{r}-\mathbf{r}_q\left(t\right)$
The first partial derivative is easy:
$\frac{\partial \left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)^i}{\partial t}=\frac{d \left(-\mathbf{r}_q\left(t\right)\right)^i}{d t}=-\mathbf{v}^i$
for the second one note that, for any function
$f=f\left(\mathbf{r}-\mathbf{r}_q\right)$
we have
$\frac{\partial f\left(\mathbf{r}-\mathbf{r}_q\right)}{\partial\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right)^i}=\frac{\partial f\left(\mathbf{r}-\mathbf{r}_q\right)}{\partial \mathbf{r}^i}=\boldsymbol{\nabla}_i f\left(\mathbf{r}-\mathbf{r}_q\right)$.
Now, delta function is not a normal function so taking derivatives of it may create problems, and should instead be handeled using a test function and an integral, but in this case such problems do not arise, so we proceed.
$\dot{\rho}\left(\mathbf{r},\,t\right)=-q \mathbf{v}^i \boldsymbol{\nabla}_i \delta^{(3)}\left(\mathbf{r}-\mathbf{r}_q\left(t\right)\right) = - \boldsymbol{\nabla}.\left(q\mathbf{v} \delta^{(3)}\left(\mathbf{r}-\mathbf{r}_q\right)\right)$
The last step is allowed since velocity does not depend on position (only on time).
Now finally you need to find a current density such that $\dot{\rho}+\boldsymbol{\nabla}.\mathbf{J}=0$
The solution you are trying to achieve is clearly a valid solution given the shape of $\dot{\rho}$. It is not, however unique, and there is not much you can do about it. Some may appeal to Special Relativity to establish uniqueness, but I have usually seen the charge and current density of a single charge being used in prooving that four-current was a four-vector, so that would be a circular argument.
following the comment.
Firstly, a disclaimer. I am not an expert on the theory of delta and other generalized functions, but I have used them a lot, primarily in electromagnetism, so do not trust me blindly, but here it goes.
Basic hand-wavey definition of the delta function would be a function that is zero everywhere except at one point AND the integral of that function is 1. More info here https://en.wikipedia.org/wiki/Dirac_delta_function. Much more info here:
M. J. Lightill, "An Introduction to Fourier Analysis and Generalized Functions" 1964
M. Rahman, "Applications of Fourier Transforms to Generalized Functions" 2011
Delta functions are not continuos so some of the normal notions you have for other functions may not work. The two problems I have seen in my practice are:
(1) Products of delta functions. What is the value of $\delta\left(x-a\right)\delta\left(x-b\right)$? Well, it is not defined. One way to think about a delta-function is that it is a limit of a very narrow Gaussian normalized to give unit integral. The key-word here is LIMIT. Two delta-functions: two limits,you need to define which limit is taken first, or how do these limits relate to each other.
For me, generalized functions are only 'safe' when they are inside an integral. For example $\delta\left(x-a\right)$ is zero everyhere and infinite at $x=a$. But stick it inside the integral with another 'good' function $f$ (continuous, differentialble, square-integrable, see books) and you get:
$\int dx \delta\left(x-a\right)f\left(x\right)=f\left(a\right)$.
(2) Derivatives of delta functions. Derivatives of delta functions make no sense on the face of it since delta functions are either zero or discontinuos, but again, stick it inside the integral:
$\frac{df}{dx}=\lim_{\epsilon\to 0}\frac{f\left(x+\epsilon/2\right)-f\left(x-\epsilon/2\right)}{\epsilon}=\lim_{\epsilon\to 0}\frac{\int dx' \delta\left(x'-\left(x+\epsilon/2\right)\right)f\left(x'\right)-\int dx' \delta\left(x'-\left(x-\epsilon/2\right)\right)f\left(x'\right)}{\epsilon}$
Here the left-hand side certainly makes sense, so we can define the right-handside, with the derivative of the delta function, with a proviso that the limit on the delta-functions should only be taken once they sit in the integral, then:
$\frac{df}{dx} = \int dx' \frac{d\delta\left(x'-x\right)}{dx}f\left(x'\right)=\int dx' \frac{d\left(x'-x\right)}{dx}\frac{d\delta\left(x'-x\right)}{d\left(x'-x\right)}f\left(x'\right)=-\int dx' \delta'\left(x'-x\right) f\left(x'\right)$
That's probably enough. In my main topic I have mentioned that derivatives of delta functions should not give you problems. The reason for this is that the most likely case where you will need the charge current (e.g.) is $\int_V d^3 r' G\left(\mathbf{r}-\mathbf{r'}\right)\rho\left(\mathbf{r'}\right)$, where $G$ is the Greens function and the observer ($\mathbf{r}$) is located outside of $V$. In this case $G$ is well-behaved and the trick with doign derivative of the delta-function once it is inside the integral, will work. If the observer is inside $G$, then the Green's function is not a 'good' function, and a lot more care is needed. For example see: W. C. Chew, "Waves and Fields in Inhomogeneous Media" 1995
Finally, there is an excellent paper by C. P. Frahm, "Some novel delta-function identities", Am. J. Phys. 51, 826 (1983). It really helped me to get friendy with these beasts.
Best Answer
No, a steady-state current is not constant throughout space, it is only constant in time. You are correct that $\vec\nabla\cdot \vec J = 0$ implies $\partial_t\rho = 0$ by the continuity equation, but that the charge inside arbitrary volumes doesn't change doesn't mean something about the current "behind & in-front" of the volume.
It means the same amount of charge flows into every volume as flows out of it, and this is precisely the natural language equivalent of $\vec\nabla\cdot\vec J = 0$, since $\vec J$ is the flow of charge, and the divergence measures the net change of charge within a volume by the continuity equation/Stokes' theorem. But that the net flow out of every volume is zero doesn't mean the vector field is constant everywhere, since every vector field that is a curl of another is divergence-less, hence an admissible steady-state current, but surely not every curl is constant throughout space.