I would say no; an example is when $R$ is a hyperbolic plane, the relation you demand says merely that $Q_1$ and $Q_2$ are in the same genus. This is in SPLAG, page 378 in the first (1988) edition, see also [Clark Jagy][1].
Notice that rational equivalence "without essential denominator" is exactly how Siegel defined the genus. So, $x^2 + 14 y^2$ and $2 x^2 + 7 y^2$ are rationally equivalent, and we can arrange that the denominators are prime to $14,$ but **not integrally equivalent**.
Give me a few minutes, I am going to display integer equivalence for $Q_1 = x^2 + 14 y^2,$ $Q_2 = 2 x^2 + 7 y^2,$ which pair take up a good deal of room in Cox's book, and $R = 2 z w.$ By solving $P^T A_1P = A_2$ in 4 by 4 integer matrices...
$$
A_1 =
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 14 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
$$
$$
P =
\left(
\begin{array}{rrrr}
2 & 7 & -2 & 2 \\
1 & 5 & -2 & 1 \\
-4 & -14 & 6 & -3 \\
2 & 14 & -5 & 3
\end{array}
\right)
$$
$$
A_2 =
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 & 7 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
$$
$$
\small
\left(
\begin{array}{rrrr}
2 & 1 & -4 & 2 \\
7 & 5 & -14 & 14 \\
-2 & -2 & 6 & -5 \\
2 & 1 & -3 & 3
\end{array}
\right)
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
0 & 14 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
\left(
\begin{array}{rrrr}
2 & 7 & -2 & 2 \\
1 & 5 & -2 & 1 \\
-4 & -14 & 6 & -3 \\
2 & 14 & -5 & 3
\end{array}
\right) =
\left(
\begin{array}{rrrr}
2 & 0 & 0 & 0 \\
0 & 7 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0
\end{array}
\right)
$$
check with gp-pari
=====================
parisize = 4000000, primelimit = 500509
? a1 = [ 1,0,0,0; 0,14,0,0; 0,0,0,1; 0,0,1,0]
%1 =
[1 0 0 0]
[0 14 0 0]
[0 0 0 1]
[0 0 1 0]
? a2 = [ 2,0,0,0; 0,7,0,0; 0,0,0,1; 0,0,1,0]
%2 =
[2 0 0 0]
[0 7 0 0]
[0 0 0 1]
[0 0 1 0]
? p = [ 2,7,-2,2; 1,5,-2,1; -4,-14,6,-3; 2,14,-5,3]
%3 =
[2 7 -2 2]
[1 5 -2 1]
[-4 -14 6 -3]
[2 14 -5 3]
? matdet(p)
%4 = 1
? pt = mattranspose(p)
%5 =
[2 1 -4 2]
[7 5 -14 14]
[-2 -2 6 -5]
[2 1 -3 3]
? a1
%6 =
[1 0 0 0]
[0 14 0 0]
[0 0 0 1]
[0 0 1 0]
? pt * a1 * p
%7 =
[2 0 0 0]
[0 7 0 0]
[0 0 0 1]
[0 0 1 0]
? a2
%8 =
[2 0 0 0]
[0 7 0 0]
[0 0 0 1]
[0 0 1 0]
? pt * a1 * p - a2
%9 =
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
[0 0 0 0]
?
===================
[1]: http://alpha.math.uga.edu/~pete/Clark_Jagy_11_13_2013.pdf