No, in general $\alpha$ doesnt generate a normal basis. The smallest counterexample is $p=5$, $r=3$. In fact for all $p\geq 5$, $r=(p+1)/2$ gives a counterexample: For $1\leq i\leq p-1$, let $\varphi_i\in \text{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$ be the automorphism given by $\zeta\mapsto \zeta^i$. A small computation shows that $\varphi_i(\alpha)+\varphi_{p-i}(\alpha)=1$ for all $i$. In particular we have $\alpha-\varphi_2(\alpha)-\varphi_{p-2}(\alpha)+\varphi_{p-1}(\alpha)=0$ and hence the conjugates of $\alpha$ are not linearly independent for $p\geq 5$.
There are other counterexamples as well: For $p=11$ the counterexamples are given by $r=3$, $r=6$ and $r=9$.