This does indeed work because this construction works for any matrix $H$ where $H^2=\mathbb{I}$.
$$
R_\theta(H)=e^{-i\theta H/2}=\cos\frac{\theta}{2}\mathbb{I}-i\sin\frac{\theta}{2}H
$$
There are several ways that you could prove this. I think the easiest way is to realise that because $H^2=\mathbb{I}$, then the eigenvalues of $H^2$ are 1, and hence the eigenvalues of $H$ must be $\pm 1$. Let $P_{\pm}$ be projectors onto the eigenspaces of eigenvalue $\pm 1$ respectively. So,
$$
\mathbb{I}=P_++P_-\qquad H=P_+-P_-
$$
Furthermore, by definition of how a function can be applied to a matrix (you apply the function to each of the eigenvalues),
$$
R_\theta(H)=e^{-i\theta/2}P_++e^{i\theta/2}P_-.
$$
Now we can substitute for $P_{\pm}$:
$$
R_\theta(H)=e^{-i\theta/2}(\mathbb{I}+H)/2+e^{i\theta/2}(\mathbb{I}-H)/2
$$
Group the terms of $\mathbb{I}$ and for $H$ and simplify.