If you are asking about 1) the usual Euclidean metric on $\mathbb{R}^3$ restricted to the (upper sheet of the) hyperboloid versus 2) the constant negative curvature metric (i.e. the hyperbolic metric) on that hyperboloid, which is obtained by restricting the Minkowski metric of $\mathbb{R}^{2,1}$ on it, then the answer is no, the hyperboloid model is not conformal. The angles measured between two hyperbolic geodesics with respect to 1) are in general different from angles measured with respect to 2). The only point of conformality (where the two angle measurements always coincide) is the apex of the sheet. Occasionally, there could be some coincidence for some very special cases, but generically the angles are different.
I forgot the bonus question: the circles in the hyperboloid model are intersection of the hyperboloid sheet with space-like planes, not passing through the origin. In particular, if you choose a point on the hyperboloid, which means you have a Minkowski-unit vector from the origin to the chosen point, you can take the space-like plane Minkowski-orthogonal to that vector and passing through the point. The latter plane is tangent to the hyperboloid. By taking planes, parallel to that one and shifted above it so that they intersect the hyperboloid, you get the family of concentric circles centered at the chosen point. Similarly, horocycles are the planes that are paralle to a light-like vector (null-vector) and offset from the origin, so that when they intersect the hyperboloid, a parabola is formed.