We claim that the answer is no.

We provide a construction where the in-sphere and circum-sphere's centres coincide. Consider two rectangles of side lengths $6$ and $8$ with overlapping centres, positioned at $90$ degrees of each other. Then, lift one into the air at a distance of $4\sqrt{3}$, such that the two rectangles are still parallel and the line connecting their centres is perpendicular to both rectangles. We claim that this cuboid satisfies the desired. Let $O$ be the midpoint of the line connecting the centres. It's distance to each of the vertices is $\sqrt{(2\sqrt{3})^2+3^2+4^2}=\sqrt{37}$, so $O$ is the circumcenter of the cuboid. We now notice that the cuboid's face are two rectangles with circumradius $\frac{1}{2}\sqrt{6^2+8^2}=5$ and of six isosceles trapezoids. Each of the trapezoids have top base $6$, bottom base $8$, height $\sqrt{\left(\frac{8-6}{2}\right)^2+(4\sqrt{3})^2}=7$. It is then easy to verify that its circumradius is also $5$. Thus, each of the faces cuts a circle of radius $5$ out of its circumsphere, implying that the perpendicular distance from $O$ is equidistant to each circumcenter of the faces. It follows that $O$ is also the incenter.