The answer as to the surjectivity of $\alpha$ is no. As in algebraic number theory, the simplest way to prove that an element is not a norm is by local considerations. Let us consider $$ y=\frac{(x^3-3x-1)(x^3+3x^2-1)}{x^2(x+1)^2}, $$ and ask whether $y$ is a norm of $\mathbb{Q}(x)$. Let us suppose that it is, and derive a contradiction. If $y$ is a norm, we would have $$ y = f(x)f(\phi(x))f(\phi^2(x))f(\psi(x))f(\psi(\phi(x)))f(\psi(\phi^2(x))), $$$$ y = \prod_{\sigma \in G} f(\sigma(x)) = f(x)f(\phi(x))f(\phi^2(x))f(\psi(x))f(\psi(\phi(x)))f(\psi(\phi^2(x))), $$ for some rational function $f \in \mathbb{Q}(x)$, where $G \subset \operatorname{Aut}(\mathbb{Q}(x))$ and $\phi,\psi \in \operatorname{Aut}(\mathbb{Q}(x))$ are as you defined them.
Now, since $g=x^3-3x-1$ is an irreducible factor of the numerator of $y$, it must appear as a numerator in at least one of the six factors in the above product representation of $y$ as well (when written "in lowest terms", i.e. after cancelling any common factors of numerator and denominator). As can be easily checked, $\phi$ permutes the roots of $g$, which means $g$ must appear in either three or all six of the factors. Letting $\xi$ be one of the zeros of $g$, this would imply that the order of the zero of the function $y$ (say considered as a meromorphic function in the variable $x$) at $x=\xi$ is a multiple of three. However from the formula for $y$ it is clear that the order of the zero at $x=\xi$ equals $1$, contradiction.