Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.

For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$. Then, the Bocher integral $\int_0^T X(t)\\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\\,dt$ for almost every $\omega$.

First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that $$ \int_0^T\lVert X(t)-\xi_n(t)\rVert_p\\,dt\le2^{-n}. $$ The Bochner integral is given by $$ \int_0^T\xi_n(t)\\,dt \rightarrow\text{(B-)}\int_0^T X(t)\\,dt. $$ Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.

Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\\,dt\right] &=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\\,dt\cr &\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\\,dt\cr &\le\sum_{n=1}^\infty2^{-n}=1 < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.

For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$. Then, the Bocher integral $\int_0^T X(t)\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\,dt$ for almost every $\omega$.

First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that $$ \int_0^T\lVert X(t)-\xi_n(t)\rVert_p\,dt\le2^{-n}. $$ The Bochner integral is given by $$ \int_0^T\xi_n(t)\,dt \rightarrow\text{(B-)}\int_0^T X(t)\,dt. $$ Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.

Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt\right] &=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\,dt\cr &\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\,dt\cr &\le\sum_{n=1}^\infty2^{-n}=1 < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. First by (uniform) continuity in the $L^2$ norm, for each fixed $T > 0$ and integer $n > 0$, there exists a $\delta_n > 0$ such that $$ \mathbb{E}\left[(X(t)-X(s))^2\right] \le 2^{-n} $$ for all $0\le s\le t\le T$ with $\lvert s-t\rvert\le\delta_n$. Choose partitions $0=t^n_0\le t^n_1\le\cdots\le t^n_{k_n}=T$ with mesh $\max_i(t^n_i-t^n_{i-1})\le\delta_n$. As in the question, define the simple functions $$ \xi_n(t) = \sum_{i=0}^{k_n-1}X(t_i^n)1_{\lbrace t_i^n \le t < t_{i+1}^n\rbrace} $$ so that, again as mentioned in the question, the Bochner integral is given by $$ \int_0^T\xi_n(t)\\,dt \rightarrow\text{(B-)}\int_0^T X(t)\\,dt. $$ Here the limit is taken in the $L^2$ norm and, hence, also holds for convergence in probability. However, each $t\in[0,T)$, lies in an interval $[t^n_i,t^n_{i+1})$ with $t-t^n_i\le\delta_n$, so $$ \mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]=\mathbb{E}\left[\lvert X(t)-X(t^n_i)\rvert\right]\le2^{-n/2}. $$ Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\lvert X(t)-\xi_n(t)\rvert\\,dt\right] &=\int_0^T\sum_{n=1}^\infty\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\\,dt\cr &\le\int_0^T\sum_{n=1}^\infty2^{-n/2}\\,dt\cr &=T/(2^{1/2}-1) < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\lvert X(t)-\xi_n(t)\rvert\\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question.

For a probability space $(\Omega,\mathcal{F},\mathbb{P})$, let $X\colon[0,T]\to L^p(\mathbb{P})$ ($1\le p\le\infty$) be Bochner integrable w.r.t the Lebesgue measure on $[0,T]$, and also jointly measurable as a map $(t,\omega)\mapsto X(t)(\omega)$ from $[0,T]\times\Omega$ to $\mathbb{R}$. Then, the Bocher integral $\int_0^T X(t)\\,dt$ agrees with the pathwise Lebesgue integral $\int_0^TX(t)(\omega)\\,dt$ for almost every $\omega$.

First, this statement clearly holds for simple functions, which are finite linear combinations of terms of the form $X(t)(\omega)=1_{\lbrace t\in A\rbrace}1_{\lbrace\omega\in B\rbrace}$, for $A$ a Borel subset of $[0,T]$ and $B$ in $\mathcal{F}$. Now, by definition, if $X$ is Bochner integrable then, for each $n\ge1$, there is a simple $\xi_n$ such that $$ \int_0^T\lVert X(t)-\xi_n(t)\rVert_p\\,dt\le2^{-n}. $$ The Bochner integral is given by $$ \int_0^T\xi_n(t)\\,dt \rightarrow\text{(B-)}\int_0^T X(t)\\,dt. $$ Here the limit is taken in the $L^p$ norm and, hence, also holds for convergence in probability.

Using pathwise Lebesgue integration along the sample paths of $X$ now, we can use Fubini's theorem to commute expectation, integration and summation signs. $$ \begin{align} \mathbb{E}\left[\int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\\,dt\right] &=\sum_{n=1}^\infty\int_0^T\mathbb{E}\left[\lvert X(t)-\xi_n(t)\rvert\right]\\,dt\cr &\le\sum_{n=1}^\infty\int_0^T\left\lVert X(t)-\xi_n(t)\right\rVert_p\\,dt\cr &\le\sum_{n=1}^\infty2^{-n}=1 < \infty. \end{align} $$ In particular, $$ \int_0^T\sum_{n=1}^\infty\left\lvert X(t)-\xi_n(t)\right\rvert\\,dt < \infty $$ with probability one. Looking at any sample path for which this is finite, $\lvert X(t)-\xi_n(t)\rvert\to0$ as $n\to\infty$ for Lebesgue almost every $t$. Also, $\lvert X(t)-\xi_n(t)\rvert$ is dominated by its sum over $n$. Therefore dominated convergence applies, $$ \int_0^T\xi_n(t)\\,dt\rightarrow\textrm{(L-)}\int_0^T X(t)\\,dt. $$ This holds for almost every sample path of $X$, so the limit holds in probability. Hence the Lebesgue integral on sample paths agrees with the Bochner integral.