Of course, it depends on the Riemannian metric you put on $\mathbb O \mathbb P^1 \cong S^8$. For the round metric you do indeed get $\mathrm{SO}(8)$ holonomy. A priori, one could imagine other "natural" metrics with reduced holonomy. However, we know by the Berger classification that if the metric was not locally symmetric, since the dimension is $8$, the only possibilities for the holonomy would imply the existence of a non-trivial parallel $2$-form or $4$-form, which is impossible in this case because of the topology of $S^8$.

I suppose that one could find some way to write $S^8 = G/H$ as a Riemannian homogenous space, in which case the holonomy would be $H$, but I am not the right person to ask about this. The only way I know how to do this is to take $G = \mathrm{SO}(9)$ and $H = \mathrm{SO}(8)$, which gives us back the round $S^8$ again.