Yes, it works in much the same way for any $k$. Here's an elementary proof.

Let $Q_k(n)$ be the number of $k$-th power free integers $\leq n$. Then $$ Q_k(n) = \sum_{d^k \leq n} \mu(d) \lfloor n/d^k \rfloor = \sum_{d^k \leq n} \mu(d) \, (n/d^k + \theta_k) $$ for some $\theta_k \in [0,1)$. Hence $$ \Bigl| \, Q_k(n) - \sum_{d^k \leq n} \frac{\mu(d)}{d^k} n \, \Bigr| < n^{1/k}. $$ But $\sum_{d^k \leq n} \mu(d)/d^k$ is a partial sum of a series that converges to $1/\zeta(k)$, with error bounded by $\sum_{d^k > n} 1/d^k \ll n^{1/k}/n$. Therefore $Q_k(n) = n/\zeta(k) + O(n^{1/k})$, QED.

Yes, it works in much the same way for any $k$. Here's an elementary proof.

Let $Q_k(n)$ be the number of $k$-th power free integers $\leq n$. Then $$ Q_k(n) = \sum_{d^k \leq n} \mu(d) \lfloor n/d^k \rfloor = \sum_{d^k \leq n} \mu(d) \, (n/d^k + \theta_d) $$ for some $\theta_d \in [0,1)$. Hence $$ \Bigl| \, Q_k(n) - \sum_{d^k \leq n} \frac{\mu(d)}{d^k} n \, \Bigr| < n^{1/k}. $$ But $\sum_{d^k \leq n} \mu(d)/d^k$ is a partial sum of a series that converges to $1/\zeta(k)$, with error bounded by $\sum_{d^k > n} 1/d^k \ll n^{1/k}/n$. Therefore $Q_k(n) = n/\zeta(k) + O(n^{1/k})$, QED.