Suppose we have a 3-dimensional space of constant, negative Gaussian curvature.

For some large enough R, construct the locus of points at distance R from a fixed center C. ... an ordinary sphere. ... Now, this sphere must itself have constant Gaussian curvature.

Yes.

Select a point P on this sphere, and begin drawing circles of increasing radius r on the sphere centered at P. I mean r to be measured along the surface of the sphere, not straight through the hyperbolic 3-space in which the sphere is embedded.

...the circumference of a circle, centered at P, changes with its radius r. For small r, it looks like 2 pi r. But as r gets larger, the negative curvature of the ambient hyperbolic space begins to make itself felt, and the circumference goes above 2 pi r...

No. But it would, if r measures the radius that goes straight through the hyperbolic 3-space.

(If it doesn't, you didn't make R big enough -- that is exactly what I meant by "large enough R" above.)

Well: if the sphere is instead a horosphere, then it's surface is Euclidean, and the circumference is 2 pi r. If a hypersphere, the surface is negatively curved, and the circumference is above 2 pi r. A horo- or hyper-sphere has constant Gaussian curvature, but isn't a locus of points at distance R from a fixed center. --- Hmm... Let theta be the angle subtended by that curved radius at the sphere center C. Then the circumference is 2 pi r sin(theta) / theta regardless of the space's (uniform) curvature. -- Don Reble djr@nk.ca