The Cayley projective plane can be realized as the compact homogeneous space $F_4/\mathrm{Spin}(9)$. In this way one can compute the curvature of this symmetric space in terms of a suitable orthonormal basis and the Lie brackets of the basic vectors. But is there any elegant expression for the curvature of $\mathbb{OP}^2$ which is independent of this homogeneous description due to the fact that $\mathbb{OP}^2$ is a rank one symmetric space?

I'm not sure what you'll count as 'elegant' or as a 'description', but here's something you may find interesting or useful: Recall that the space of curvature tensors $\mathcal{K}(V)$ on an inner product space $V$ is essentially the kernel of the natural map $S^2\bigl(\Lambda^2(V^\ast)\bigr)\to \Lambda^4(V^\ast)$. Now let $V$ be the $16$-dimensional spin representation of $\mathrm{Spin}(9)$. It is well known (and easy to see via highest weights) that one has $$ \Lambda^2(V^\ast) = \Lambda^2(W)\oplus\Lambda^3(W) $$ where $W$ is the standard $9$-dimensional representation of $\mathrm{Spin}(9)$ (which, of course, factors through $\mathrm{SO}(9)$). Each of these summands is irreducible. It follows that, as $\mathrm{Spin}(9)$-representations $$ S^2\bigl(\Lambda^2(V^\ast)\bigr) = S^2\bigl(\Lambda^2(W)\bigr)\oplus \bigl(\Lambda^2(W)\otimes\Lambda^3(W)\bigr) \oplus S^2\bigl(\Lambda^3(W)\bigr), $$ and hence it has two trivial summands, i.e., the space of curvature tensors fixed under $\mathrm{Spin}(9)$ has dimension $2$. (You don't lose either of these in the map to $\Lambda^4(V^\ast)$ since this latter space has no trivial summands under $\mathrm{Spin}(9)$.) These represent the Ricci curvature (the Cayley plane is an Einstein manifold) and the Weyl curvature of the Cayley plane.

To get explicit formulae, you just need to make the above decomposition of $\Lambda^2(V^\ast)$ explicit. The first summand is obvious, since it is the Lie algebra of $\mathrm{Spin}(9)$, for the second, you need a model of how to see the wedge product of two elements in $V$ as the sum of a $2$-form and a $3$-form on $W$. Of course, this can be done by using the Clifford algebra, which can be studied using the octonions. I think that's probably as explicit as you can make it.

I'm not quite sure what you are looking for, but explicit computation of the curvature appears in arXiv:math/0702631. The authors define $\mathbb{OP}^2$ via an octonionic atlas, write down explicit isometries with which they prove that it is indeed a homogeneous manifold and then they express the curvature tensor at a point using a coordinate frame. The expressions of the components of $R$ involve only octonionic multiplication and inner product.