Consider a rectangle $R=[a_1,b_1]\times [a_2,b_2] \times ... \times [a_n,b_n]$. For $\epsilon >0$ sufficiently small $R'=[a_1-\epsilon ,b_1+\epsilon ]\times [a_2-\epsilon ,b_2+\epsilon ] \times ... \times [a_n-\epsilon ,b_n+\epsilon ]$ contains $R$ in its interior and its volume tends to volume of $R$ as $ \epsilon \to 0$. Hence the volume of $R'$ is at most equal to $2v(R)$ for $\epsilon$ sufficiently small. .
Consider the more general case where $\int_S f = \int_Q f_S$ exists but $f$ is not everywhere continuous in $Q$. Let $D_f\subset S$ denote the set of discontinuity points of $f$ in $S$.
With $A = int(S)$, the set $D$ of discontinuity points for $f_S$ is
$$D = (A \cap D_f) \,\cup\, (\partial S \cap D_f) \,\cup\, \{x_0 \in \partial S \setminus D_f: \lim_{x \to x_o, x \in A} f(x) \neq 0\} $$
Since $f_S$ is integrable on $Q$ the set $D$ is of measure zero.
On the other hand, $f_A$ vanishes everywhere on $\partial S$ and $f_A = f_S$ for all $x \in A$. Hence, the set $E$ of discontinuity points for $f_A$ is
$$E = (A \cap D_f) \,\cup \, \{x_0 \in \partial S: \lim_{x \to x_o, x \in A} f(x) \neq 0\}$$
Note that
$$ \{x_0 \in \partial S: \lim_{x \to x_o, x \in A} f(x) \neq 0\} \subset (\partial S \cap D_f) \,\cup\, \{x_0 \in \partial S \setminus D_f: \lim_{x \to x_o, x \in A} f(x) \neq 0\} $$
Hence, $E\subset D$ and $E$ is also of measure zero.
The argument that $\int_S f = \int_A f$ remains the same. It is irrelevant if $f_S - f_A$ does not vanish at points of $D\setminus E$ since $D\setminus E \subset D$ is of measure zero.
Best Answer
As stated in the beginning, "... each of the rectangles $Q,Q_1,\ldots, Q_k$ is a union of subrectangles determined by $P$".
Thus, $Q_j = \bigcup_{l=1}^{m_j} R_{jl}$ for each $j=1,\ldots,k$ and since $Q_1,\ldots, Q_k$ cover $Q$, we have
$$Q \subset \bigcup_{j=1}^k Q_j = \bigcup_{j=1}^k\bigcup_{l=1}^{m_j}R_{jl}$$
If $R \subset Q$, then as a member of the partition $P$ it must belong to the set $\{R_{jl}\}$ and so is contained in at least one of the rectangles $Q_1, \ldots , Q_k$.