Let $r=\alpha+\beta$. The condition of the problem guarantees that $r\in\mathbb Q$. Besides, the definition of $r$ yields $\beta=r-\alpha$, so that $\mathbb Q\left(\beta\right) = \mathbb Q\left(r-\alpha\right) = \mathbb Q\left(\alpha\right)$. The minimal polynomial of $\alpha$ is $P$ (since $\alpha$ is a root of $P$ and since $P$ is irreducible). Thus, there exists a $\mathbb Q$-algebra isomorphism $\mathbb Q\left[X\right] / \left(P\right) \to \mathbb Q\left(\alpha\right)$ which sends $\overline R$ to $R\left(\alpha\right)$ for every polynomial $R \in \mathbb Q\left[X\right]$. This isomorphism clearly sends the residue class $\overline{Q\left(r-X\right)}$ to $Q\left(\underbrace{r-\alpha}_{=\beta}\right)=Q\left(\beta\right)=0$. Hence, the residue class $\overline{Q\left(r-X\right)}$ must itself be $0$ (because if an isomorphism sends some element to $0$, then the element must itself be $0$). In other words, $Q\left(r-X\right)\equiv 0\mod \left(P\right)$. In yet other words, $P\mid Q\left(r-X\right)$. Thus, $\deg P\leq \deg \left(Q\left(r-X\right)\right)=\deg Q$. Similarly, $\deg Q\leq \deg P$. Combining these two inequalities, we get $\deg P=\deg Q$. Thus, $\deg P=\deg Q=\deg\left(Q\left(r-X\right)\right)$. Hence, $P\mid Q\left(r-X\right)$ can only be possible if $Q\left(r-X\right)=sP$ for some scalar $s\in\mathbb Q$. Consider this $s$. Since $P$ is monic, the leading coefficient of $sP$ is $s$. Since $Q$ is monic, the leading coefficient of $Q\left(r-X\right)$ is $\pm 1$ (actually, $\left(-1\right)^{\deg P}$, to be precise). Hence, comparing the leading coefficients in the equality $Q\left(r-X\right)=sP$, we obtain $\pm 1=s$. Thus, $Q\left(r-X\right)=\pm P$. Applying this to $X=\frac{r}{2}$, we get $Q\left(\frac{r}{2}\right)=\pm P\left(\frac{r}{2}\right)$, so that $Q\left(\frac{r}{2}\right)^2=P\left(\frac{r}{2}\right)^2$. Hence, the polynomial $P^2-Q^2$ has a rational root, namely $\frac{r}{2}$.

Here's another proof of $Q(r-X) = \pm P(X)$ in language a little friendlier. Essentially this is a proof that minimal polynomials are unique. Suppose $f \in \mathbb{Q} [X]$ has $\alpha$ as a root. I claim $P(X) \mid f(X)$. If not, then since $P$ is irreducible we have $\gcd(P, f) = 1$. By Bezout, we can find $r, s \in \mathbb{Q} [X]$ such that \[rP + sf =1.\]This is clearly bad since $\alpha$ is a root of the left side but not the right, so the claim is proved. Now, returning to the problem at hand, note that $Q(r-X)$ has $\alpha$ as a root, so $P(X) \mid Q(r-X)$. But $Q$ is irreducible, so $Q(r-X)$ is irreducible; hence $\deg Q = \deg P$. As $ P, Q$ are monic this gives $Q(r-X) = \pm P(X)$ (with sign depending on the parity of $\deg Q$), as desired.