No. (The main idea here is present in Dylan Wilson's comment.)

Every principal $SU(2)$-bundle over $S^2$ is trivial, because $\pi_1 SU(2)$ is trivial. But there is a nontrivial oriented bundle over $S^2$ with fiber $S^3$, namely the unit sphere bundle of the nontrivial rank $4$ vector bundle. (There are precisely two rank $r$ oriented vector bundles over $S^2$ if $r\ge 3$, because $\pi_1 SO(r)$ has order two.)

I should explain why the nontrivial vector bundle has nontrivial unit sphere bundle, that is, why $\pi_1 SO(4)$ injects into $\pi_1 SDiff(S^3)$.

You can use the big theorem of Hatcher ("Smale Conjecture"), which says that $\pi_k SO(4)$ maps isomorphically to $\pi_k SDiff(S^3)$ for all $k$.

Alternatively, you can use that the bundle is detected by the Stiefel-Whitney class $w_2$, and that Stiefel-Whitney classes of vector bundles are invariant under fiber homotopy equivalence of unit sphere bundle.

Or you can use $\pi_3$ instead of $\pi_1$ as suggested by Dylan Wilson, making a bundle over $S^4$; there are elements of $\pi_3 SO(4)\cong \mathbb Z\times \mathbb Z$ not coming from $\pi_3 SU(2)\cong\mathbb Z$. Again, the resulting sphere bundle is nontrivial either by Hatcher's theorem or by using characteristic classes. I'm not sure what the most elementary version of the characteristic-class argument would be.

One example of a rank $4$ real vector bundle over $S^4$ that does not admit a complex structure is the tangent bundle!