Note that, in general, every axiom of $T$ is equivalent to the quantifier-free formula $0=0$ under the definition you have given. So for the first question it is sufficient to find any axiom that is not logically equivalent to a $\Sigma^0_1$ formula.
There is such an axiom for $Q$, namely $(\forall y)(y = 0 \lor (\exists x)(y = S(x)))$. This axiom is $\Pi^0_2$ and it cannot be logically equivalent to a $\Sigma^0_1$ formula because it is not preserved under taking superstructures.
The fact in my first paragraph also gives information on the second question, because it means that any arithmetical formula that is true in $\mathbb{N}$ is $\Sigma^0_1$ over $\operatorname{Th}(\mathbb{N})$. So all you need for the second question is a formula true in $\mathbb{N}$ but not equivalent to a $\Sigma^0_1$ formula over $Q$.
Furthermore, because $Q$ proves any true $\Sigma^0_1$ formula, it is sufficient to find any arithmetical formula that is true in $\mathbb{N}$ but not provable in $Q$. Because if a true formula $\phi$ was provably equivalent over $Q$ to a $\Sigma^0_1$ formula, then because the axioms of $Q$ are true the $\Sigma^0_1$ formula would be true, hence provable in $Q$, and thus $\phi$ would also be provable in $Q$. One concrete example of a true formula not provable in $Q$ is the Paris--Harrington principle.