There's some confusion about the question, so let me try answering a similar question that I think is strictly analogous to yours but easier. If you can confirm that the question I'm asking addressing really is analogous to yours, and that the answer is the kind of answer you're looking for, then maybe that will help prepare the ground for someone to answer your actual question.
So, let's pretend that you were interested in a much more rigid kind of geometry, so that a triangle consists of three joined-up straight line segments. There will be no curvature involved.
Formally, we might work with affine spaces, and define a triangle in an affine space $A$ to be an ordered triple of points in $A$. Of course, we imagine the three points joined up, but that won't be part of the formalism.
A generic triangle is an affine space $P$ equipped with a triangle $(p_1, p_2, p_3)$, with the property that for any affine space $A$ and triangle $(a_1, a_2, a_3)$ in $A$, there is a unique affine map $f: P \to A$ such that $$ (f(p_1), f(p_2), f(p_3)) = (a_1, a_2, a_3). $$ The affine space $P$ is what you call the classifying space in your question.
It's easy to see that there does exist a generic triangle (and by the universal property, it's unique up to isomorphism). Indeed, we can take $P$ to be any two-dimensional affine space and $p_1, p_2, p_3$ to be any three affinely independent points of $P$.
Of course you know all this, Six Winged Seraph. But the point of saying it is to ask: is this the type of answer you want, in the more advanced setting that you describe?