Integral in different coordinate systems In Griffiths' electrodynamics book, he uses the equation,
$$\nabla^2\mathbf{A}=-\mu_0 \mathbf{J},$$
to state that
$$\mathbf{A}(\mathbf{r}) = \frac{\mu_0}{4\pi}\int\frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}'-\mathbf{r}|}\mathrm{d}\tau'.$$
This is, of course, justified by the fact that each cartesian component of $\mathbf{A}$ obeys Poisson's equation, according to the first equation.
But then he went on to say that to evaluate the integral you are restricted to use cartesian coordinates because that was our assumption in deriving the second equation from the first. (4th edition, page 244, footnote 19).
This seems wrong to me. As far as I can imagine, the value of the integral is independent of the system of coordinates you use. 
 A: Your intuition is correct that you should be able to use this formula in any coordinate system. Indeed, you have presumably done problems with the scalar potential where you have a similar formula that you can evaluate in any coordinate system. However you have to be more careful when defining what you mean by this integral in the vector case in general coordinate systems, and I think Griffiths wants to avoid getting mired in those subtleties so he issues a warning that is perhaps a bit too strongly worded.
But let's plunge ahead anyway. Starting with the equation you wrote
\begin{equation}
\mathbf{A}(\mathbf{r})=\int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d \tau'
\end{equation}
This is not super useful as it stands, because it involves the integral of a vector. In practice you always end up computing integrals of scalars. So the first step is to convert this vector integral into three scalar integrals.
For example, one of the scalar integrals is
\begin{equation}
A_x = \mathbf{A}(\mathbf{r})\cdot\hat{\mathbf{e}}_x=\hat{\mathbf{e}}_x\cdot\int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d \tau' = \int \frac{\mathbf{J}(\mathbf{r}')\cdot{\hat{\mathbf{e}}_x}}{|\mathbf{r}-\mathbf{r}'|}d \tau'= \int \frac{J_x(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d \tau'
\end{equation}
The key step is the last one, which looks extremely innocent but is actually quite tricky. The point here is that the unit vector $\mathbf{\hat{e}}_x$ is a constant, so even though we started off by evaluating it at the observer coordinate $\mathbf{r}$, we are free to "move" the unit vector to the source coordinate $\mathbf{r}'$, and thus we can easily perform the dot product between the unit vector and the current. This is the step that gets complicated if you try to do things in other coordinate systems. 
As an example where things get hairy, let's try to evaluate things in spherical coordinates, so we want to compute $A_r,A_\theta,A_\phi$.
So try computing $A_r$. We need $\hat{\mathbf{e}}_r$ so we can dot it with $\mathbf{A}$. However, $\hat{\mathbf{e}}_r$ is not a constant. It's magnitude is always 1, but its direction depends on where you are in space (explicitly, $\hat{\mathbf{e}}_r(\mathbf{R})=\frac{\mathbf{R}}{|\mathbf{R}|}$). We want to evaluate $A_r$ at the observer coordinate $\mathbf{r}$, so similarly we want to evaluate $\hat{\mathbf{e}}_r$ at the point $\mathbf{r}$.
With this in mind, we run through the steps we did in Cartesian space
\begin{equation}
A_r = \mathbf{A}(\mathbf{r})\cdot\hat{\mathbf{e}}_r(\mathbf{r})=\hat{\mathbf{e}}_r(\mathbf{r})\cdot\int \frac{\mathbf{J}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d \tau' = \int \frac{\mathbf{J}(\mathbf{r}')\cdot{\hat{\mathbf{e}}_r(\mathbf{r})}}{|\mathbf{r}-\mathbf{r}'|}d \tau'
\end{equation}
But now
\begin{equation}
\mathbf{J}(\mathbf{r}')\cdot{\hat{\mathbf{e}}_r}(\mathbf{r})\neq J_r(\mathbf{r'})
\end{equation}
Instead, you have to write
\begin{equation}
\mathbf{J}(\mathbf{r}')\cdot\hat{\mathbf{e}}_r(\mathbf{r}) = J_r(\mathbf{r}') \hat{\mathbf{e}}_r(\mathbf{r'})\cdot\hat{\mathbf{e}}_r(\mathbf{r}) +  J_\theta(\mathbf{r}') \hat{\mathbf{e}}_\theta(\mathbf{r'})\cdot\hat{\mathbf{e}}_r(\mathbf{r})+ J_\phi(\mathbf{r}') \hat{\mathbf{e}}_\phi(\mathbf{r'})\cdot\hat{\mathbf{e}}_r(\mathbf{r})
\end{equation}
Hopefully you can see this is going to become a real pain!
You can, of course, evaluate the dot products explicitly using geometry (for example, $\hat{\mathbf{e}}_r(\mathbf{r})\cdot\hat{\mathbf{e}}_r(\mathbf{r}')=\cos\theta$ if $\phi=\phi'$), and you will be left with a sum of three integrals to do (note you now need to do three integrals to evaluate $A_r$, one of the three components of $\mathbf{A}$!). There are some tricks you can try to use; for example you can pick your source coordinates such that the integrals you need to do are as nice as possible, and you also have some gauge freedom you can try to use to help simplify things.
However, I would never suggest actually doing any of the above in practice (unless you are in a situation with a lot of symmetry where you can reduce all the dot products to something very simple). I am just trying to be explicit to illustrate what is going on. At the level of Griffiths, your best bet is generally to compute the Cartesian components of the vector potential $A_x,A_y,A_z$. If you ever need the radial component $A_r$, you can construct it from $A_x,A_y,A_z$ in the usual way. 
Finally, I just want to point out that while it is simplest to compute the Cartesian components $A_x,A_y,A_z$, the actual integrations for the components can be carried out in any coordinate system. In other words, when evaluating
\begin{equation}
A_x = \ \int \frac{J_x(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|}d \tau'
\end{equation}
where $J_x=\mathbf{J}(\mathbf{r}')\cdot\hat{\mathbf{e}_x}$, you are free to use any coordinates for $\mathbf{r}'$ you like to evaluate the integral. The subtleties are all about the basis vectors, once you have chosen an appropriate basis vector and formed a scalar integral, you are free to use any coordinate system you like to evaluate the integral, with no hidden subtleties over the normal scalar potential (ie, electrostatic) case.

(This was in the original answer, I moved it down here so as not to disrupt the flow of the above).
If you really want to know how to compute $\mathbf{A}(\mathbf{r})$ (or $\mathbf{E}$ and $\mathbf{B}$) in spherical coordinates, without going through Cartesians, the 'right' way to do it is in terms of the vector spherical harmonics. There is a fair amount of additional sophisticated (but interesting!) machinery you need, so explaining them is beyond the scope of this answer, but if you are interested see Jackson, 3rd edition, sections 5.6 and 9.7, or the sections on the vector multipole expansion and the inhomogeneous helmholtz equation in the lecture notes by Fitzpatrick http://farside.ph.utexas.edu/teaching/jk1/Electromagnetism/Electromagnetism.html, or the wikipedia article http://en.wikipedia.org/wiki/Vector_spherical_harmonics. 
A: The integral follows from the use of a Green's function for the vector Laplacian; that Green's function is not coordinate system dependent.  The integral can be derived in any coordinate system.
The problem here is more of a practical one:  the vector $A(r)$ can be expressed in terms of a basis of vector fields--in particular, the values of those basis fields at the point $r$.  The big issue is with the $r'$ parts of the integral: most approaches to doing integrals that result in vectors as the final answer use a fixed (e.g. cartesian) basis so that the integral breaks down into scalar integrals multiplied by fixed basis vectors.
A: I don't have this book at hand, so I don't know how it is derived there. But I guess the integral formula comes from Green's function approach to solving a Poisson equation. However, as far as I know, the Green's function method is working for the normal Poisson equation, i.e., the unknown function is a scalar instead of a vector A like here. In order to get the integral equation, we could think of the 1st equation as three Poisson equations for the Cartesian components, and assemble the solutions to the three Poisson equations into a vector again. For this purpose however coordinates other than Cartesian don't work. 
As for if there is any counterpart integral formula suitable for all coordinate systems... I look forward to other people's answers. 
