sIt seems that this argument hasn't been presented yet, so I might as well include it.

We can sort the integers $n \in [1, X]$ by their radicals, which is necessarily a square-free integer $m$. Thus we have

$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} = \sum_{\substack{m \leq X \\ m \text{ square-free}}} \frac{1}{m} \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1.$$

Now, $\text{rad}(n) = m$ if and only if $p | n \Rightarrow p | m$. If we write $m = p_1 \cdots p_k$, then

$$\displaystyle \sum_{\substack{n \leq X \\ \text{rad}(n) = m}} 1 = \#\{(x_1, \cdots, x_k) : x_i \in \mathbb{Z} \cap [0,\infty), p_1^{x_1} \cdots p_k^{x_k} \leq X/m\}.$$

The inequality defining the right hand side is equivalent to

$$\displaystyle x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m),$$

and this is just counting integer points with non-negative entries bounded by a simplex, and ~~it is easy to see that~~

$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} \ll \frac{\log(X/m)}{\prod_{1 \leq i \leq k} \log(p_i)} \ll \log X.$$

EDIT: This last step is wrong, but it can be fixed. Indeed, we can arrange the $p_i$'s so that $p_1 < p_2 < \cdots < p_k$. It then follows from Davenport's lemma that

$$\displaystyle \# \{(x_1, \cdots, x_k) : x_1 \log p_1 + \cdots + x_k \log p_k \leq \log(X/m)\} = O \left(\sum_{i=0}^k \frac{(\log X/m)^{k-i}}{\prod_{1 \leq j \leq k-i} \log p_i} \right).$$

It then follows that

$$\displaystyle \sum_{n \leq X} \frac{1}{\text{rad}(n)} \ll \sum_{\substack{p_1 < \cdots < p_k \\ p_1 \cdots p_k \leq X}} \sum_{i=0}^k \frac{(\log X)^{k-i}}{\prod_{1 \leq j \leq k -i} p_i \log p_i}.$$

From here I think it is possible to get the bound $O_\epsilon(X^\epsilon)$, but it requires a somewhat more refined analysis on the interaction between the number of primes and the size of the primes.