No; one way this can become impossible is if the first $k-1$ paths "use up" all the vertices that a $k^{\text{th}}$ path would need to use. For example:
![enter image description here](
This graph is $2$-connected, and there are $2$ internally disjoint paths from the left vertex to the right vertex. But if you start with the path highlighted in red, then you can't find a second path internally disjoint from the red path.