By the uniform continuity theorem, $f$ is uniformly continuous on $Q$.
Thus, given $\varepsilon>0$, there exists $\delta>0$ such that when $\mathbf x,\mathbf y \in Q$ and $||\mathbf x-\mathbf y||<\delta$, then $|f(\mathbf x)-f(\mathbf y)|<\varepsilon/v(Q)$.
Let $m$ be so large that the $||\mathbf a-\mathbf b||$ divided by $m$ is less than $\delta$ and consider the mth regular partition of $Q\,$ (defined in an obvious way).
Then, using the extreme value theorem, you have$$\left|\sum_Rv(R)(M_R(f)-m_R(f))\right|<\epsilon,$$ where $R$ ranges over all the subrectangles of the partition and $M_R(f)$ is the supremum and $m_R(f)$ the infimum of $f$ over $R$.
Therefore, by the Riemann criterion, $f$ is integrable over $Q$.
Define the width of $Q = [a_1,b_1] \times [a_2,b_2] \times \ldots \times [a_n,b_n]$ to be $W_Q = \max_{1\leqslant j \leqslant n}(b_j - a_j)$. Let $N$ be the number of subrectangles in the partition $P_0$ and take $\delta = \epsilon/(4MN W_Q^{n-1})$ where $|f(x)| \leqslant M$ for all $x \in Q$.
If $\|P \| \leqslant \delta$ and $P'= P \cup P_0$ is the common refinement, then
$$\tag{*}U(f,P) \leqslant U(f,P') + N \cdot 2M \cdot W_Q^{n-1} \cdot \delta \leqslant U(f,P') + \frac{\epsilon}{2}$$
We have $U(f,P') \leqslant U(f,P_0)$ since $P'$ is a refinement of $P_0$, and it follows that
$$U(f,P) \leqslant U(f,P_0) + \frac{\epsilon}{2} \leqslant \overline{\int_Q} f + \epsilon$$
To understand the bound, the partition $P'$ has at most (largely overestimating) $N$ more subintervals of $[a_1,b_1]$ than the partition $P$ and the width of these subintervals is bounded by $\delta$. Each of these extra subintervals $[\alpha_{1j},\beta_{1j}]$ is the edge of multiple rectangles in $P'$ extending in the other dimensions. One such rectangle contributes to the difference $U(f,P) - U(f,P’)$ by no more than the maximum oscillation $2M$ times the volume of the rectangle. The total volume of the rectangles with edge $[\alpha_{1j},\beta_{1j}]$ is conservatively bounded above by $W_Q^{n-1} \cdot \delta$.
Elaboration on first inequality in (*)
Consider a slice $\mathcal{S}$ of $P$-rectangles $R_1,\ldots,R_m$ which can be written as $R_j = [\alpha_{1j},\beta_{1j}]\times T_j$ where the intervals $[\alpha_{1j},\beta_{1j}]$ form a partition of $[a_1,b_1]$ and the $T_j$ are $(n-1)$-dimensional rectangles. Each rectangle $R_j$ is a union of rectangles $R_{jk}\subset R_j$ from the refined partition $P'$.
Let $M_j = \sup_{x \in R_j}f(x)$ and $M_{jk} = \sup_{x \in R_{jk}} f(x)$. Since $|f(x)| \leqslant M$, we have $M_j < M_{jk} + 2M$ (although there is always one rectangle $R_{jk}$ with $M_{jk} = M_j$).
With $R_{jk} = [\alpha_{1,jk}, \beta_{1,jk}] \times T_{jk}$ we have $vol(R_{jk}) \leqslant \delta \,vol(T_{jk})$. In forming the refinement $P'$ by merging $P_0$ and $P$ we create no more than $N$ rectangles in $P'$ where the supremum of $f$ does not coincide with the containing rectangle in $P$.
Thus,
$$\sum_{R_j \in \mathcal{S}}M_j \, vol(R_j) \leqslant \sum_{R_j \in \mathcal{S}}\sum_{R_{jk} \subset R_j} M_{jk} \, vol(R_{jk}) +N \cdot 2M \cdot \delta \cdot \max_{R_{jk} \subset R_j \in \mathcal{S}} vol(T_{jk})$$
Summing over all slices $\mathcal{S}$ of $Q$ we recover the upper sums
$$U(f,P)= \sum_{\mathcal{S}}\sum_{R_j \in \mathcal{S}}M_j \, vol(R_j), \,\, U(f,P') = \sum_{\mathcal{S}}\sum_{R_j \in \mathcal{S}}\sum_{R_{jk} \subset R_j} M_{jk}\, vol(R_{jk}), $$
and obtain the inequality
$$U(f,P) \leqslant U(f,P') + N \cdot 2M \cdot \delta \cdot \sum_{\mathcal{S}}\max_{R_{jk} \subset R_j \in \mathcal{S}} vol(T_{jk}) \\ \leqslant U(f,P') + N \cdot 2M \cdot \delta \cdot W_Q^{n-1} $$
Best Answer
$\nu(f,a) < \epsilon$ means that there exists $\delta(a) > 0$ such that $\sup \{ \lvert f(x) - f(y) \rvert : x,y \in B(a;\delta(a)) \cap I \} < \epsilon$.
Since $\nu(f,a) < \epsilon$ for all $a \in I$, we conclude that if a subrectangle $R \subset I$ is contained in some $B(a;\delta(a)) \cap I$, then $M_R(f) -m_R(f) < \epsilon$.
$\mathcal{B} = \{ B(a;\delta(a)) \cap I : a \in I \}$ is an open cover of $I$. There exists a Lebesgue number $\rho > 0$ for $\mathcal{B}$. This means that that every subset of $I$ having diameter less than $\rho$ is contained in some member of $\mathcal{B}$.
Now choose a partition $P$ of $I$ such that all elements have diameter less than $\rho$.
Edited:
I add a proof for the existence of a Lebesgue number. I shall do this in a form specialized to your situation.
The diameter of a bounded subset $M \subset \mathbb{R}^n$ is defined as
$$diam(M) = \sup \{ \lvert x - y \rvert : x,y \in M \} .$$
We claim that there exists $\rho > 0$ such that for all $M \subset I$ with $diam(M) < \rho$ there exists $a \in I$ such that $M \subset B(a;\delta(a)) \cap I$.
If our claim would not be true, then we could find a sequence of nonempty subsets $M_n \subset I$ with $diam(M_n) < \frac{1}{n}$ which are contained in no $B(a;\delta(a))$. Choose $x_n \in M_n$. Then $(x_n)$ is a sequence in $I$ and must have a subsequence converging to some $x \in I$ because $I$ is compact. Therefore we can find a sufficiently large $n$ such that both $diam(M_n) < \frac{1}{2}\delta(x)$ and $\lvert x_n - x \rvert < \frac{1}{2}\delta(x)$. Hence for $y \in M_n$ $$\lvert y - x \rvert \le \lvert y - x_n \rvert + \lvert x_n - x \rvert < \delta(x)$$ which means $M_n \subset B(x;\delta(x)$, a contradiction.