[First paragraph has been edited, after Vitali's comments below.] According to one convention, hyperKahler manifolds are *not* actually quaternionic-Kahler. This is the case if you define hyperKahler as having holonomy exactly $\mathrm{Sp}(m)$, and quaternionic-Kahler as having holonomy exactly $\mathrm{Sp}(n) \cdot \mathrm{Sp}{1} = (\mathrm{Sp}(n) \times \mathrm{Sp}(1)) / \lbrace \pm 1 \rbrace$, where in both cases the real dimension is $4m$.

In fact, these "strictly" quaternionic-Kahler manifolds are NOT even Kahler, so they are poorly named, but unfortunately, the name has stuck. HyperKahler manifolds are always Ricci-flat, but quaternionic-Kahler manifolds are always Einstein with nonzero Einstein constant.

By contrast, hypercomplex manifolds need not have a Riemannian metric at all. Being hypercomplex means that one has a triple $I, J, K$ of almost complex structures which satisfy the quaternion multiplication relations. I can't remember if one requires them to all be integrable or if this is always true. On a hyperKahler manifold, we certainly have such a structure, so it is hypercomplex. A quaternionic-Kahler manifold does not admit such complex structures *globally*, only *locally*, but the $4$-form $\Omega = \omega_I^2 + \omega_J^2 + \omega_K^2$ is actually globally defined, where $\omega_I = g(I \cdot, \cdot)$ is the associated Kahler form to $I$.

Added later: Quaternionic projective space $\mathbb H \mathbb P^n$ is *not* hyperKahler. For example, $\mathbb H \mathbb P^1 \cong S^4$, which cannot even be Kahler, since $b_2(S^4) = 0$.