$\newcommand{\E}{\operatorname{\mathsf{E}}}$ You do not need the separability of $B$ to define $\E F$ for a random vector $F\colon\Omega\to B$; however, you need to assume that $F$ is strongly measurable, in the sense that there is a sequence of finitely-valued random vectors $F_n$ in $B$ such that $\|F_n(\omega)-F(\omega)\|\to0$ for almost all $\omega\in\Omega$.
By Bochner's theorem, if $F$ is strongly measurable and $\E\|F\|<\infty$, then $F$ is Bochner-integrable, in the sense that for some sequence of finitely-valued random vectors $F_n$ in $B$ we have $\|F_n(\omega)-F(\omega)\|\to0$ for almost all $\omega$ and $\E\|F_n-F\|\to0$; then $\E F:=\lim_n EF_n$, with naturally defined $\E F_n$.
It is then a known and easy to check fact that $\E RF=R\E F$ for any strongly measurable random vector $F$ with $\E\|F\|<\infty$ and any bounded linear operator $R\colon B\to\mathbb R$; that is, the Bochner integrability implies the Pettis integrability.
See e.g. Yosida, Section V.4 for further details.
Concerning your remark that usually the Banach space is assumed to be separable: this is done to ensure the measurability in a number of instances, including the measurability of the sum of random vectors.