Addressing the second question: It seems to me that if you have an embedding of $S^{n-1}\times I$ into $\mathbb R^n$, this is the same as an ambient isotopy of one copy of $S^{n-1}$ to another. (I found this easier to visualize when thinking of the fact that you can't embed a cylinder in $\mathbb R^2$ except as an annulus of concentric circles.)

In particular, you're trying to get one sphere to bound the same disk as the other sphere does, by a homotopy whose image at any time never intersects the image at another time. You've already noted that two concentric spheres bound the same disk: so you're trying to get the first sphere $A$ to surround the other sphere $B$. Since $B$ cuts off a component of $\mathbb R^n$, we can remove that component and know the isotopy will not pass through it without first passing through $B$, which is not allowed. Therefore remove a ball $C$ or a point from the interior of $B$. $A$ is contractible in $\mathbb R^n-C$, and $B$ is not: $B$ generates the $n-1$st homology group of the resulting manifold. This also is a nice way to see why it will work in $\mathbb R^{n+1}$, though of course in that case you really do have a "cylinder" (take $I$ to run in the $n+1$st dimension with each $S^{n-1}$ in an $n$-hyperplane).

Edit: This is meant to answer the question of why we can't have an embedding $\mathbb S^{n-1}\times I\hookrightarrow\mathbb R^n$ such that the boundary is two side-by-side spheres rather than two nested spheres. I said "the second question" but it changed.

It seems to me that if you have an embedding of $S^{n-1}\times I$ into $\mathbb R^n$, this is the same as an ambient isotopy of one copy of $S^{n-1}$ to another. (I found this easier to visualize when thinking of the fact that you can't embed a cylinder in $\mathbb R^2$ except as an annulus of concentric circles.)

In particular, you're trying to get one sphere to bound the same disk as the other sphere does, by a homotopy whose image at any time never intersects the image at another time. You've already noted that two concentric spheres bound the same disk: so you're trying to get the first sphere $A$ to surround the other sphere $B$. Since $B$ cuts off a component of $\mathbb R^n$, we can remove that component and know the isotopy will not pass through it without first passing through $B$, which is not allowed. Therefore remove a ball $C$ or a point from the interior of $B$. $A$ is contractible in $\mathbb R^n-C$, and $B$ is not: $B$ generates the $n-1$st homology group of the resulting manifold. This also is a nice way to see why it will work in $\mathbb R^{n+1}$, though of course in that case you really do have a "cylinder" (take $I$ to run in the $n+1$st dimension with each $S^{n-1}$ in an $n$-hyperplane).