As Tsemo already answered, this is false.
But now you might wonder if it's still true that $f(x^2)$ is irreducible if we are given any irreducible $f(x) \in \mathbb Q[x]$ with $\deg(f) > 1$. Now, if $\alpha \in \mathbb Q$ is a root of $f(x^2)$, then $\alpha^2 \in \mathbb Q$ is a root of $f(x)$ and hence $(x - \alpha^2)$ is a factor of $f(x)$. This means if $f(x)$ is not linear then the degree of the factors of $f(x^2)$ must be at least $2$. So I looked for examples with $\deg(f) = 2$ with $f(x^2)$ having two quadratic factors. Remembering a factoring trick, I found the following example.
Take $f(x) = x^2 + 4 \in \mathbb Q[x]$ which is irreducible since it has no roots in $\mathbb Q$. Then we have,
f(x^2) = x^4 + 4 = x^4 + 4x^2 + 4 - 4x^2 = (x^2 + 2)^2 - (2x)^2 = (x^2 - 2x + 2)(x^2 + 2x + 2)
and so $f(x^2)$ is not irreducible.