No.
Let $(X, \cal A, \mu)$ be $[0,1]$ with Lebesgue measure.
Let $E = L^2[0,1]$ with inner product $\langle \alpha,\beta\rangle := \int \alpha(t)\overline{\beta(t)}\;dt$.
Define $f : [0,1] \to L^2[0,1]$ by
$$
f(t) = \mathbf1_{[0,t]}
$$
(I wrote $\mathbf1_S$ for the indicator function of a set $S$.)
Now $f$ is continuous and bounded, so it is Bochner integrable.
The Bochner integral $\psi= \int_0^1 f(t)\;dt$ satisfies: for all $\alpha \in L^2$,
$$
\langle \psi, \alpha\rangle
=\int_0^1 \psi(s) \overline{\alpha(s)}\;ds
=\int_0^1 \int_0^1\mathbf1_{[0,t]}(s)\;dt \;\overline{\alpha(s)}\;ds
\\
=\int_0^1 \int_s^1 1\;dt \;\overline{\alpha(s)}\;ds
=\int_0^1 (1-s)\;\overline{\alpha(s)}\;ds
$$
Therefore $\psi(s) = 1-s$ for almost all $s \in [0,1]$. (Altering $\psi$ on a nullset, we may assume $\psi(s) = 1-s$ for all $s \in [0,1]$.)
Now let us consider
$$
\varphi =\sum_{k=1}^\infty\lambda_k f(x_k) ,
$$
where $\lambda_k$ and $x_k$ are as specified.
Can we have $\varphi(s) = 1-s$?
Arguing as before, we conclude
$$
\varphi(s) = \sum_{k=1}^\infty\lambda_k \mathbf1_{[0,x_k]}(s)
= \sum_{x_k \ge s} \lambda_k
$$
for almost all $s$. (Again we may assume this holds for all $s$.)
Because the $\lambda_k$ are nonnegative, we see that $\varphi$ is a nonincreasing function. We may assume without loss of generality that the points $x_k$ are all distinct. Because $\varphi(s) = 1-s$, for some $k$ we have $0 < x_k < 1$ and $\lambda_k > 0$.
By re-labeling, we may assume $0<x_1<1$ and $\lambda_1 > 0$. Now compute
$$
\lim_{s \nearrow x_1} \varphi(s) = \sum_{x_k \ge x_1}\lambda_k,
\qquad
\lim_{s \searrow x_1} \varphi(s) = \sum_{x_k > x_1}\lambda_k .
$$
The function $\varphi$ has a jump of size $\lambda_1 > 0$ at $x_1$.
Recalling that $\varphi$ is nonincreasing, we conclude that $\varphi(s) = 1-s$ cannot hold almost everywhere.
Thus $\varphi \ne \psi$.
