Taking the Poincaré disk model for hyperbolic space as being formed by all complex numbers with modulus smaller than 1, the distance between two points can be calculated as
$d(z_1,z_2)=\tanh^{-1}\left|\frac{z_1-z_2}{1-z_1\bar{z_2}}\right| \; .$
The set of points at equal distance $\varepsilon=\tanh(d(z,z_0))$ from a fixed point $z_0$ can be shown to be represented by a circle in the Poincaré disk model with center
$\frac{(1-\varepsilon^2)z_0}{1-\varepsilon^2|z_0|^2}$
and radius
$\frac{\varepsilon(1-|z_0|^2)}{1-\varepsilon^2|z_0|^2} \; .$
But this is not the end of the story, you can't just pick points uniformly on this circle. What you want is to pick points such that you can arrive from $z_0$ to them such that the direction of departure from $z_0$ was chosen uniformly. This will induce a particular distribution on that circle.
To figure out this distribution, it is necessary to compute the intersection of an h-line* through $z_0$ for an arbitrary direction and the circle we just computed. From that, we can get the sought distribution. I'll update this later if I get to computing that part.
EDIT: Following the suggestion by @Douglas Zare, this can also be achieved by picking a point at a distance $\varepsilon$ from the origin uniformly, where the deformation of the Poincaré disk model won't be a problem since h-lines passing through the origin are all just diameters. All we have to do then is translate the point with the translation that takes the origin to the point $z_0$ which can be done with a Möbius transformation
$z \mapsto w=\frac{z+z_0}{\bar{z}_0z+1} \; .$
This is also easily implemented computationally. Here's how I coded it in R:
# Brownian motion on Poincaré disk (no comments about crappy code plz kthx) epsilon=0.01; z<-0; path<-c(z); for (t in 1:10000) { jitter=runif(1,0,1); dz=epsilon*complex(1,cos(2*pi*jitter),sin(2*pi*jitter)); z<-(z+dz)/(Conj(dz)*z+1); path<-c(path,z); } plot(Re(path),Im(path),type="l",col="blue",asp=1,xlim=c(-1,1),ylim=c(-1,1)); curve(sqrt(1-x^2),-1,1,col="red",add=TRUE); curve(-sqrt(1-x^2),-1,1,col="red",add=TRUE);
And here is a picture of a run of the code 
*By h-line, I mean a line in the hyperbolic plane, which will be either a diameter, either a circle segment orthogonal on the boundary of the Poincaré disk.