Under reasonable assumptions about $\Sigma$ the answer is yes. For example if $\Sigma$ is smooth and compact $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ and it has trivial normal bundle*, that follows from the tubular neighborhood theorem. According to this theorem a neighborhood $U\subset\mathbb{R}^n$ of $\Sigma$ is diffeomorphic to $B^m\times\Sigma$. If $F:U\to B^m\times\Sigma$ is the diffeomorphism, $\pi:B^m\times\Sigma\to B^m$ is the projection onto the first factor, and $f=\pi\circ F$, then for every $y\in B^m$, $f^{-1}(y)$ is diffeomorphic to $\Sigma$.

This should also be true for smooth non-compact submanifolds with trivial normal bundle, because you can make tubular neighborhood gradually decrease as you move away. However, I did not think about that case carefully.

*There was a mistake in my previous answer and as observed by Andy Putman in his comment.

Under reasonable assumptions about $\Sigma$ the answer is yes. For example if $\Sigma$ is smooth and compact $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ and it has trivial normal bundle*, that follows from the tubular neighborhood theorem. According to this theorem a neighborhood $U\subset\mathbb{R}^n$ of $\Sigma$ is diffeomorphic to $B^m\times\Sigma$. If $F:U\to B^m\times\Sigma$ is the diffeomorphism, $\pi:B^m\times\Sigma\to B^m$ is the projection onto the first factor, and $f=\pi\circ F$, then for every $y\in B^m$, $f^{-1}(y)$ is diffeomorphic to $\Sigma$.

This should also be true for smooth non-compact submanifolds with trivial normal bundle, because you can make tubular neighborhood gradually decrease as you move away. However, I did not think about that case carefully.

*There was a mistake in my previous answer and as observed by Andy Putman in his comment.

Under reasonable assumptions about $\Sigma$ the answer is yes. For example if $\Sigma$ is smooth and compact $(n-m)$-dimensional submanifold of $\mathbb{R}^n$, that follows from the tubular neighborhood theorem. According to this theorem a neighborhood $U\subset\mathbb{R}^n$ of $\Sigma$ is diffeomorphic to $B^m\times\Sigma$. If $F:U\to B^m\times\Sigma$ is the diffeomorphism, $\pi:B^m\times\Sigma\to B^m$ is the projection onto the first factor, and $f=\pi\circ F$, then for every $y\in B^m$, $f^{-1}(y)$ is diffeomorphic to $\Sigma$.

This should also be true for any smooth non-compact submanifolds because you can make tubular neighborhood gradually decrease as you move away. However, I did not think about that case carefully.

Under reasonable assumptions about $\Sigma$ the answer is yes. For example if $\Sigma$ is smooth and compact $(n-m)$-dimensional submanifold of $\mathbb{R}^n$ and it has trivial normal bundle*, that follows from the tubular neighborhood theorem. According to this theorem a neighborhood $U\subset\mathbb{R}^n$ of $\Sigma$ is diffeomorphic to $B^m\times\Sigma$. If $F:U\to B^m\times\Sigma$ is the diffeomorphism, $\pi:B^m\times\Sigma\to B^m$ is the projection onto the first factor, and $f=\pi\circ F$, then for every $y\in B^m$, $f^{-1}(y)$ is diffeomorphic to $\Sigma$.

This should also be true for smooth non-compact submanifolds with trivial normal bundle, because you can make tubular neighborhood gradually decrease as you move away. However, I did not think about that case carefully.

*There was a mistake in my previous answer and as observed by Andy Putman in his comment.