Unless I am misunderstanding something, the following seems to be a counterexample.
Given the standard Euclidean norm in $\mathbb{R}^3$ and the plane $x_1=.6$, $H\cap\partial C$ is the circle $ \{x:x_1=.6\text{ and }x_2^2+x_3^2=.64\} $ No matter what $z$ is (other than $0$), the elliptic cylinder $(H \cap \partial C) + \mathbb{R}z$ will intersect $\mathring{C}$.
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Given any $z$, if there is an $h\in H\cap\partial C$ so that $h+\mathbb{R}z$ does not intersect $\mathring{C}$, then $z$ must be perpendicular to the normal to $\partial C$ at $h$. In order that $h+\mathbb{R}z$ does not intersect $\mathring{C}$ for all $h\in H\cap\partial C$, $z$ would have to be perpendicular to each normal to $\partial C$ at a point of $H\cap\partial C$.
Therefore, $z$ would need to be perpendicular to $(.6,.8,0)$ and $(.6,0,.8)$, thus parallel to their cross product $(.64,-.48,-.48)$. Furthermore, $z$ would need to be perpendicular to $(.6,-.8,0)$ and $(.6,0,.8)$, thus parallel to their cross product $(-.64,-.48,.48)$. However, since $(.64,-.48,-.48)$ and $(-.64,-.48,.48)$ are not parallel, $z$ cannot be parallel to them both, unless it is $0$.
Thus, there is no $z$ that so that $\left[(H\cap\partial C)+\mathbb{R}z\right]\cap\mathring{C}=\emptyset$.
Revised Question
Consider the maximum norm on $\mathbb{R}^3$ given by $ \|x\|=\max(|x_1|,|x_2|,|x_3|) $ Under this norm, $C$ is a cube of side $2$.
Let $H$ be the plane $x_1+x_2+x_3=0$. This plane intersects all $6$ sides of $\partial C$. Each of the following points is on $H$ and a different side of $\partial C$: $ \small\left\{\left(1,-\frac12,-\frac12\right),\left(-\frac12,1,-\frac12\right),\left(-\frac12,-\frac12,1\right),\left(-1,\frac12,\frac12\right),\left(\frac12,-1,\frac12\right),\left(\frac12,\frac12,-1\right)\right\} $
As argued in the answer to the original question, $z$ must be perpendicular to the normal to $C$ at all points of $H\cap\partial C$. However, the normals to the cube span $\mathbb{R}^3$, so there can be no such $z$ other than $z=0$.
True in Two
Due to the fact that $\|-x\|=\|x\|$, the normal to $\partial C=\{x:\|x\|=1\}$ at any $x$ is the opposite of the normal at $-x$. In $\mathbb{R}^2$, $H\cap\partial C$ consists of two points, $h$ and $-h$. Thus, if $z$ is perpendicular to the normal to $\partial C$ at $h$, then neither $h+\mathbb{R}z$ nor $-h+\mathbb{R}z$ intersect $\mathring{C}$.