No. The reason this holds in the complex case (any degree $0$ map $S^1 \to S^1$ factors through $\exp: \mathbb{R}i\to S^1$) is that in that case the exponential map is a covering map, in particular a local homeomorphism. This tells you that there is never a local problem when constructing a lift, and you only need things to be compatible globally, which is where the degree condition comes in.

In the quaternion case, the exponential map is not a local homeomorphism: it collapses the sphere of radius $\pi k$ in $\Im \mathbb{H}$ to the element $(-1)^k\in S^3$, for each $k$. Now take a geodesic path $I\subseteq S^3$ joining $-1$ and $1$. There exists a degree $0$ map $f: S^3 \to S^3$ which is the identity in an open neighbourhood of $I$ (by adding suitable local degrees away from $I$). I claim this map cannot lift through $\Im \mathbb{H}$. Indeed, if it does, it would need to take $-1$ and $1$ to spheres of radius $k\pi$ and $(k+1)\pi$ in $\Im\mathbb{H}$ for some $k\in \mathbb{Z}$, by looking at how the lift needs to vary along $I$. But then at least one of $k,k+1$ is nonzero, say $k$. Now I claim there is simply no section to the map $\exp: \Im\mathbb{H}\to S^3$, restricted to a neighbourhood of the sphere of radius $k\pi$ on the left and a neighbourhood of $(-1)^k$ on the right. Indeed, a continuous section would take $(-1)^k$ to a single point in the sphere of radius $k\pi$, but since the (restricted) map is a homeomorphism away from the sphere of radius $k\pi$, looking at a sequence converging to any other point contradicts continuity of the section.