Of course this can happen, but generally this is not the case. As Jason suggested, the issue comes down to splitting fields. First, note that if $X$ and $Y$ are smooth projective varieties over a field $k$, and $X \dashrightarrow Y$ is a rational map, then for a field extension $L/F$, $X(L) \neq \emptyset$ implies that $Y(L) \neq \emptyset$. This can be proved by induction on the dimension of $X$, and can be reduced to the case that $L = k$ (by extending scalars). The induction case follows by blowing up a rational point on $X$, and noting that one has a rational map from the exceptional divisor, which has smaller dimension.
In particular, it follows that if $X$ and $Y$ are birational, then they have points over the same set of fields. Now, if we have a birational isomorphism $Q \sim X_1 \times \cdots \times X_r$, for Brauer-Severi varieties $X_i$ and a quadric $Q$, it follows that since $Q$ has points in quadratic extensions, each $X_i$ does as well. Of course, at least one of the $X_i$ must have no rational points, since if they all did, so would $Q$. Therefore we have a map
$Q \to X$ for some nontrivial Brauer-Severi $X$. Writing $X = BS(A)$ for a central simple algebra $A$, we find that since $X$ has a point after a quadratic extension, $A$ must be index $2$. One can then define a rational projection $X \dashrightarrow C$ where $C$ is the associated Brauer-Severi conic curve for the underlying quaternion division algebra for $A$.
The existence of the rational map $Q \dashrightarrow C$ obtained by composition will give the contradiction in general.
Consider, for example an anisotropic Pfister quadric associated to a quadratic form of dimension at least 8. For example, a dimension 8 example would look like:
$q = x_0^2 + a x_1^2 + b x_2^2 + c x_3^2 + ab x_4^2 + ac x_5^2 + bc x_6^2 + abc x_7^2$.
Such that $q$ has no nontrivial zeros over the ground field $k$. This could be arranged, for example, by making the variables $a, b, c$ indeterminates.
The associated quadric (vanishing set) $Q = X(q)$, has the property that it is "strongly 2-incompressible," (for the definition of this see http://www.math.jussieu.fr/~karpenko/publ/icm-r.pdf or http://www.math.ucla.edu/~merkurev/papers/survey-update3.pdf). This implies that if one has a rational map $Q \dashrightarrow Y$ with $\dim Y < \dim Q$, then $ind_2(Y) < ind_2(Q)$. Here, $ind_2$ is the largest power of $2$ dividing the degree of any closed point. But since $Q$ has degree $2$ closed points (by intersection with lines), it follows that $Y$ would have to have a point over some odd degree extension. Considering $Y = C$ above gives the contradiction.