With the definitions in the OP, this is false. It is OK if the Banach space $B$ is separable and $(\Omega,\mathcal F, P)$ is an arbitrary probability space. It is OK if the Banach space $B$ is arbitrary and $(\Omega,\mathcal F,P)$ is a perfect measure space. But for arbitrary $B$ and $(\Omega,\mathcal F, P)$, it can fail. It can fail in many different ways.

(A theorem of Charles Stegall: if $(\Omega,\mathcal F,P)$ is a perfect probability space, $B$ is a metric space, and $f : \Omega \to B$ is $(\mathcal F, \mathcal B)$-measurable, then there is a set $\Omega_1 \subseteq \Omega$ of measure $1$ such that $f(\Omega_1)$ is separable.)

Here is the simplest way in which it may fail. Write $\mathcal B = \mathrm{Borel}(B)$. Let $L^p(\Omega,B)$ be the set of all functions $f : \Omega \to B$ such that $f$ is $(\mathcal F, \mathcal B)$-measurable, and
$$
\int_\Omega \|f(\omega)\|^p\;dP(\omega) < \infty .
$$

It is possible that there are $f,g \in L^p(\Omega,B)$ such that $f+g \notin L^p(\Omega,B)$ because $f+g$ is not even $(\mathcal F , \mathcal B)$-measurable.

**Example I**

Let $T$ be a discrete space with cardinal $\frak{a} > 2^{\aleph_0}$. Let $B = l^2(T)$, that is, a Hilbert space with orthonormal basis of cardinal $\frak{a}$. For each $t \in T$ let $e_t \in l^2(T)$ be defined by: $e_t(t) = 1$ and $e_t(s) = 0$ if $t\ne s$. This system of "unit vectors" is an orthonormal basis of the space $B$.

Let $\Omega = T \times T$ be the Cartesian square. Let $\mathrm{Borel}(T)$ be the Borel sigma-algebra on $T$, which is of course the power set of $T$.
Let the sigma-algebra $\mathcal{F} = \mathrm{Borel}(T) \otimes \mathrm{Borel}(T)$, the product sigma-algebra. The reason for requiring that $\mathrm{card}(T) > 2^{\aleph_0}$ is so that the diagonal
$$
\Delta := \{(t,t) \in \Omega : t \in T\},
$$
although closed, is not in $\mathcal F$. See HERE.

We do not care what the probability measure $P$ is. (In an extreme case it could even be the point mass at a single point.)

Finally we are ready. Define $f : \Omega \to B$ by
$$
f\big((u,v)\big) = e_u,
$$
That is: Given $\omega = (u,v)$ in $\Omega$, we take its first component, and use the corresponding unit vector. Similarly, define $g : \Omega \to B$ by
$$
g\big((u,v)\big) = -e_v,
$$
using the second component and a minus sign.

I claim that $f, g \in L^p(\Omega,B)$ but $f+g$ is not.

First: $f$ is $(\mathcal F, \mathcal B )$-measurable. Indeed, if
$Q \in B$ is Borel, then $f^{-1}(Q) \in \mathcal F$ because
$f^{-1}(Q) = \widetilde{Q} \times T \in \mathcal F$ where
$\widetilde{Q} = \{t \in T : e_t \in Q\}$.
So $f$ is $(\mathcal F, \mathcal B )$-measurable. Similarly
$g$ is $(\mathcal F, \mathcal B )$-measurable.

Next,
$$
\int_\Omega \|f(\omega)\|^p\,dP(\omega) = 1 < \infty.
$$
(Regardless of what the probability measure $P$ is, the integral of the constant $1$ is $1$.)
So $f \in L_p(\Omega,B)$. Similarly, $g \in L_p(\Omega,B)$.

Now we claim the sum $f+g$ is not measurable. Indeed, even more, we claim that $\{\omega\in \Omega : f(\omega)+g(\omega) = 0\} \notin\mathcal F$. (Since $\{0\}$ is closed, this shows $f+g$ is not measurable.) Indeed,
$$
\{\omega : f(\omega) + g(\omega) = 0\} =
\{(u,v) : e_u-e_v = 0\} =
\{(u,v) : u=v\} = \Delta.
$$
As noted above, $\Delta \notin \mathcal F$

**End of Example I**