Reposting comment as an answer and expanding.
The answer is yes. You can find an exposition in Condensed Matter Field Theory by Altland and Simons, starting on page 134 in the second edition. The troubles come from that spin can't be described with a Hamiltonian that is a function of $q$:s and their conjugate $p$:s. However the more general formulation of Hamiltonian mechanics in terms of symplectic manifolds permits a description of spin. Altland and Simons cite Arnold's Mathematical Methods of Classical Mechanics as a reference for this. It's an under-appreciated gem of a book.
So, when we construct the path integral we consider the paths to be paths in phase space, $p$:s and $q$:s. I think to understand this in geometrical terms we have to back to "basics". The Lagrange formulation we have coordinates $q$ and velocities $\dot q$. The coordinates can be coordinates on any manifold -- this is why the Lagrangian formalism is so neat for constrained systems -- so the velocities are really tangent vectors. Lagrangian mechanics is thus naturally formulated on tangent bundles. But we can take the Legendre transform and pass to the Hamiltonian formalism with $p$:s and $q$:s. This takes us to the cotangent bundle, for $$p_i = \frac{\partial L}{\partial \dot q^i}$$
is the 1-form that is $\partial L /\partial \dot q^i$ on the vector field $\dot q^i$ and 0 on the other coordinate vector fields.
However you can do all of Hamiltonian mechanics on any manifold that comes with a symplectic structure. A symplectic structure is 2-form $\omega$ such that $d\omega = 0$ and for every vector $v$, $\omega(v, \cdot)$ is not the zero 1-form. (You can think of $\omega$ as a sort of anti-symmetric metric.) This is what Arnold does in his book. The cotangent bundle naturally comes with such an $\omega$: $$\omega = dp_1 \wedge dq_1 + \ldots dp_n \wedge dq_n$$
(this 2-form is independent of your choice of coordinates $q$). The Hamiltonian is a function on $M$, so in the particular case of a cotangent bundle, you can take it to be a function of the $p$:s and $q$:s.
Now you can couple orbital angular momentum to the vector potential just fine with $p$:s and $q$:s. But how to do it with intrinsic angular momentum, that is spin? We want a Hamiltonian like $$H = \mathbf B \cdot \mathbf S.$$
Since the spin of a particle has a definite magnitude, the dynamical part of the spin is its direction. So this Hamiltonian is defined on a manifold like $T^* M \times S^2$ (I put the first factor there because $\mathbf B$ could of course vary in space). Of course this rests on $S^2$ having an admissible $\omega$, but you can take the volume form $$\omega = \sin\theta\; d\theta \wedge d \varphi$$
with $\theta, \varphi$ the usual coordinates.
Therefore the paths you use in your construction of the path integral for spin should be paths on the sphere, and you can use $\theta, \varphi$ as coordinates. There is a formal complication in the quantum case since our states also have phases. This means that we should really use paths in $SU(2)$, since an arbitrary spin state can always be written like $$|g\rangle := g\vert\uparrow \rangle$$ where $\vert\uparrow\rangle$ is some reference state. So the resolution of identity inserted to construct the path integral should be $$\operatorname{id} \int_{SU(2)} \vert g \rangle \langle g\vert$$ (the integration is with respect to an $SU(2)$-invariant volume form.) This leads to an extra term, the usual $\partial_\tau$, in the path integral. However it turns out that the phase is irrelevant to this term, so the path integral is really over paths on the sphere $S^2$. You can find the detailed calculation in Altland and Simons.
