I would say no; thean 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.
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]
?
===================