If my calculation isn’t wrong, we can further prove that $2^{n+1} \nmid f(n)$ where $f(n)$ is the number of partitions.

Let $f(n)$ denote the number of partitions of the strip. Place the strip on the Cartesian plane with coordinates of corners $(0,0)$, $(n,0)$, $(n,1)$ and $(0,1)$. We induct on $n$ to show that $2^n | f(n)$. For $n=1$, we have $f(1)=2$: [asy][asy] size(8cm); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,0)--cycle); draw((3,0)--(4,0)--(4,1)--(3,1)--(3,0)--cycle); draw((4,1)--(3,0)); [/asy][/asy] Now consider some partition of a $1\times (n-1)$ strip. Consider moving the upper edge of the partition to the right (in a sense, applying a shear force). This means that there are no vertical segments in this new partition of the $1\times n$ strip, and also we have the option of drawing two more edges, $(0,0)-(1,1)$ and $(n-1,0)-(n,1)$. Thus we have four partitions of the $1\times n$ strip without vertical segments for each partition of the $1\times (n-1)$ strip. By the induction hypothesis, we have $2^n | 4f(n-1)$. We now count the the number of partitions where the leftmost vertical segment has $x$-coordinate equal to $k$. This segment divides the $1\times n$ strip into a $1\times k$ strip without vertical segments, and a $1\times (n-k)$ strip that can be partitioned any way. Analogously, we may show that the number of partitions is $4f(k-1)f(n-k)$ for $k>1$. When $k=1$, the number of such partitions is $2f(n-1)$. Thus, for each $k$, we get a number of partitions divisible by $2^n$.