It is.
For the sake of contradiction, suppose there is a linear combination of elements in the set over $\mathbb{Q}$ that equal 0. We have a sum of the form
$\sum_1^n \frac{q_j}{a_j-\pi}=0$
where $q_j, a_j \in \mathbb{Q}$, the $a_j$ are distinct, and $q_j\neq 0$ for all $j$. Consider the following expression:
$\sum_1^n \frac{q_j}{a_j-x}=0.$
Our problem is equivalent to showing that $\pi$ is not a root of this.
Clearing denominators yields
$\sum_{j=1}^n {q_j}\prod_{i\neq j}(a_i-x)=0.$
This is clearly a polynomial with rational coefficients. The only way $\pi$ can be a root of this is if the polynomial is the zero polynomial, as $\pi$ is a transcendental number. But if this this the zero polynomial, then setting $x=a_1$ yields
$q_1\prod_{i\neq 1} (a_i-a_1)=0$
which is a contradiction as all terms are nonzero.