The answer seems to be negative, already for $n=4$ and $n=8$.

For $n=1, 2, 4, 8$, we know there exists an identity

$$(x_1^2+\cdots+x_n^2)(y_1^2+\cdots+y_n^2)=z_1^2+\cdots+z_n^2,$$
where each $z_i$'s are bilinear on $x$'s and $y$'s, i.e., there exists
an invertible rational matrix $P$ such that $z=x^tPy$, where $x^t=[x_1\cdots
x_n], y^t=[y_1\cdots
y_n], z^t=[z_1\cdots
z_n]$.
These identities comes from the norm of real, complex, Hamilton and
Cayley numbers.
Moreover one can always choose $z_1=x_1y_1+\cdots+x_ny_n$.
Conversely Adolf Hurwitz proved that if such a identity happens then
$n=1, 2, 4, 8$.

Now for your question choose the polynomials $g_1,\cdots,g_{n-1}\in\mathbb{R}[x_1,\cdots,x_n]$ such
that $$x_1g_1+\cdots+x_{n-1}g_{n-1}=0, \ (*)$$ also put $g_n=0$.
Using the above identity one obtains
$$(x_1^2+\cdots+x_n^2)(g_1^2+\cdots+g_n^2)=z_2^2+\cdots+z_n^2.$$

Also for $n\geqslant 4$, in $(*)$ it is possible to choose $g_i$'s in such a way
that $q=g_1^2+\cdots+g_{n-1}^2$ is not divisible by $p=x_1^2+\cdots+x_n^2$.
So $m\leqslant n-1$.