inverse element in Paillier cryptosystem
Yes. $r^n$ needs to be coprime with $n^2$. The only elements which have don't an inverse modulo $p^2 q^2$ are all multiples of $p$ and all multiples of $q$, so we just require $\gcd{(r^n, p)} = \gcd{(r^n, q)} = 1$.

$\implies \gcd{(r^n, n}) = 1$

Clearly, if $r$ is coprime to $n$, then $r \times r \times \cdots \times r ~ (n ~ \mathrm{times})$ will also be coprime to $n$, so:

$\gcd{(r, n)} = 1$

The probability of a random $r \in \mathbb{Z}_n^*$ not satisfying the above is equal to $\frac{n - 1 - \varphi{(n)}}{n - 1} \to 0$.

Assuming you use sufficiently large $n$, the probability of selecting a bad $r$ (one that cannot be inverted, according to your criterion) is smaller than the probability of your hardware failing and screwing up the calculation while you getting hit by a dozen lightning bolts simultaneously. I think you'll be fine.