Of course that is true for any field $k$ where all such quadrics have $k$-points, e.g., finite fields. However, I expect that it is generally false.

Assume that $X$ is birational to a product $P$ of Severi-Brauer varieties (including, possibly $\mathbb{P}^1$ factors). If $k$ is infinite, then for a general $2$-plane section $C$ of $X$, $C$ would map into each Severi-Brauer factor of $P$. Using the Esnault-Levine-Wittenberg indices as in Kollár's recent article, this forces each Severi-Brauer factor to be a curve. So $P$ is a product of copies of $C$ and $\mathbb{P}^1$. In particular, this means that the splitting fields of $X$ are precisely the splitting fields of $C$. But I expect this is "typically" false, i.e., there should be many more splitting fields for $X$ than for $C$.

For instance, if $X$ is given by the diagonal quadratic form, $$ Q(x_0,x_1,\dots,x_n) = x_0^2 + t_1 x_1^2 + \dots + t_n x_n^2,$$ over $\mathbb{C}(t_1,\dots,t_n)$, and if the $2$-plane is the locus where $x_i=a_ix_2$ for $i>3$ and given $a_3,\dots,a_n\in \mathbb{C}$, then the splitting fields of $$ G(x_0,x_1,x_2) = x_0^2 + t_1x_1^2 + (t_2 + a_3^2t_3 + \dots + a_n^2t_n)x_2^2$$ should be far fewer than splitting fields of $Q$. For instance, for every $i=1,\dots,n$, the field extension $k(\sqrt{-t_i})$ is a splitting field of $Q$, but I do not see why that should be a splitting field for $C$ if $i>1$.

Of course that is true for any field $k$ where all such quadrics have $k$-points, e.g., finite fields. However, I expect that it is generally false.

Assume that $X$ is birational to a product $P$ of Severi-Brauer varieties (including, possibly $\mathbb{P}^1$ factors). If $k$ is infinite, then for a general $2$-plane section $C$ of $X$, $C$ would map into each Severi-Brauer factor of $P$. Using the Esnault-Levine-Wittenberg indices as in Kollár's recent article, this forces each Severi-Brauer factor to be a curve. So $P$ is a product of copies of $C$ and $\mathbb{P}^1$. In particular, this means that the splitting fields of $X$ are precisely the splitting fields of $C$. But I expect this is "typically" false, i.e., there should be many more splitting fields for $X$ than for $C$.

For instance, if $X$ is given by the diagonal quadratic form, $$ Q(x_0,x_1,\dots,x_n) = x_0^2 + t_1 x_1^2 + \dots + t_n x_n^2,$$ over $\mathbb{C}(t_1,\dots,t_n)$, and if the $2$-plane is the locus where $x_i=a_ix_2+b_ix_1$ for $i\geq 3$ and given $a_i,b_j\in \mathbb{C}$, then the splitting fields of $$ G(x_0,x_1,x_2) = x_0^2 + (t_1+b_3^2t_3+ \dots + b_n^2t_n)x_1^2 + (t_2 + a_3^2t_3 + \dots + a_n^2t_n)x_2^2$$ should be far fewer than splitting fields of $Q$. For instance, for every $i=1,\dots,n$, the field extension $k(\sqrt{-t_i})$ is a splitting field of $Q$, but I do not see why that should be a splitting field for $C$.