After looking through some literature I'm answering my own question. I'll wait on accepting for a few days in case anyone else has a better answer.
First of all, the problem is equivalent to solving,
$$A x = \sqrt{A}b,$$
so the only square root part needed is computing $\sqrt{A}b$, after which it is a normal solve.
The following paper discusses two ways to do that,
Numerical approximation of the product of the square root of a matrix with a vector,
Allen, EJ and Baglama, J and Boyd, SK ,
Linear Algebra and its Applications 2000,
http://www.sciencedirect.com/science/article/pii/S0024379500000689
The second method transforms the problem into solving an ODE whose solution at time $t=1$ is $A^{1/2}b$. This ODE is,
$$\begin{cases}
\frac{dx}{dt} = -\frac{1}{2}\left(At + (1-t)I\right)^{-1}(I-A)x(t) \\
x(0) = b,
\end{cases}$$
which can be solved through standard techniques such as forward Euler. At each timestep $t_k$, $(I-A)$ must be applied to $x(t_k)$, and then a regularized version of $A^{-1}$:
$$\left(At_k + (1-t_k)I\right)^{-1}$$
must be applied to the result. In my particular situation the method for applying $A^{-1}$ can be extended to also apply $(A + \alpha I)^{-1}$. In other situations it should be possible to use $A^{-1}$ to build a preconditioner for $\left(At_k + (1-t_k)I\right)$.
Edit: A major downside of this method is that the ODE becomes stiff as the condition number of $A$ increases.