$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ The affirmative answer to this question is provided by Scalora, Theorem 2.1, page 354, which can be stated as follows, using the setting and notation in the question:

If $F\colon\Omega\to B$ is a Bochner-integrable random vector, then there is a $\P$-almost surely unique Bochner-integrable strongly $\mathcal G$-measurable random vector $\E(F|\mathcal G)=:G$ in $B$ such that $\E G1_A=\E F1_A$ for all $A\in \mathcal G$.

Note that in Scalora's paper the strong measurability is part of the definition of the Bochner integrability. Note also that then, by Bochner's theorem, $F$ is Bochner-integrabile iff $F$ is strongly measurable and the condition $\E\|F\|<\infty$ holds.

$\newcommand{\E}{\mathsf{E}}$ $\newcommand{\P}{\mathsf{P}}$ The affirmative answer to this question is provided by Scalora, Theorem 2.1, page 354, which can be stated as follows, using the setting and notation in the question:

If $F\colon\Omega\to B$ is a Bochner-integrable random vector, then there is a $\P$-almost surely unique Bochner-integrable strongly $\mathcal G$-measurable random vector $\E(F|\mathcal G)=:G$ in $B$ such that $\E G1_A=\E F1_A$ for all $A\in \mathcal G$.

Note that in Scalora's paper the strong measurability is part of the definition of the Bochner integrability. Note also that then, by Bochner's theorem, $F$ is Bochner-integrabile iff $F$ is strongly measurable and the condition $\E\|F\|<\infty$ holds.

It is not a big advantage here to allow the Banach space to be non-separable. Indeed, the strong measurability of a random vector implies that it is almost separably valued; see e.g. the note in the middle of page 348 of Scalora's paper.