I would say no. Because a constant curvature manifold is symmetric. Any 2 dimensional subspace should be like a hyperboloid or a sphere. These objects do not have torsion.



Do exist constant curvature manifolds (hyperbolic or elliptic) with torsion?
You should consider the following example:


Let $M^3=\mathrm{SU}(2)\simeq S^3$ endowed with its biïnvariant metric $g$ (unique up to a constant multiple, let's fix this by requiring that the $g$ has constant sectional curvature equal to $1$). Now consider the unique connection $\nabla$ for which the left-invariant vector fields on $SU(2)$ are parallel. This connection is compatible with the biïnvariant metric and it has torsion, since $T(X,Y) = -[X,Y]\not=0$ when $X$ and $Y$ are linearly independent left-invariant vector fields. This metric has the same geodesics as the Levi-Civita connection, and every $2$-plane has sectional-curvature $1$.


The symmetry group of $(M^3,g,\nabla)$ is $\mathrm{SO}(4)$, the maximum possible. So this is an example of a constant curvature manifold with torsion.


As É. Cartan and J. A. Schouten showed in 1926 (*On Riemannian geometries admitting an absolute parallelism*), this example generalizes to biïnvariant metrics on compact simple Lie groups endowed with the unique connection for which the left-invariant vector fields are parallel, though they no long actually have constant sectional curvature.


However, they also show in that paper that there is a connection $\nabla$ on the $7$-sphere $S^7$ of unit octonions that is flat, compatible with the natural metric of constant curvature on $S^7$, and invariant under the group $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$. Because $\mathrm{Spin}(7)$ acts transitively on $3$-planes in $\mathbb{O}\simeq\mathbb{R}^8$, it follows that this geometry $(S^7, g,\nabla)$ also has constant curvature.


