Let $E \to M$ be a complex vector bundle over an almost complex manifold.
Is there always an almost complex structure on $E$? I.e. does there always exist some $J : E \to E$ such that $J^2 = -Id$?
My question is motivated by two cases. First, if $M$ is a complex manifold, we know this is always true. This is because on a complex manifold, almost complex structures on $E$ (where a Hermitian metric $h$ is given) are in correspondence with Hermitian/unitary connections (see here). And one can easily show that any complex bundle $(E,h)$ over $M$ admits a Hermitian connection.
Second, one can further show that any complex vector bundle over any (compact?) smooth manifold admits a Hermitian connection. This is achieved by taking the corresponding classifying map into a large enough conplex Grassmannian and pulling back the Chern connection of the tautological bundle with respect to a given metric on it. Since pullbacks of metric compatible connections are also metric compatible with respect to the pullback metric, the pullback bundle has a metric compatible connection, i.e. a Hermitian connection.
So one can take, for example, the tangent bundle of $S^4$ which we know does not admit an almost complex structure, yet it admits a Hermitian connection. (EDIT: this is not a complex vector bundle) So the correspondence between Hermitian connections and almost complex structures fails for a manifold that is not almost complex, let alone complex.
I'm wondering if this correspondence works for almost complex manifolds? I.e.:
Let $E \to M$ be a complex vector bundle over an almost complex manifold, with a Hermitian metric $h$. Are Hermitian connections on $E$ in correspondence with almost complex structures on $E$, as is the case for when $M$ is a complex manifold?