If they behave like random oracles, then they offer security commensurate with the size of the image space which is $2^\omega\binom n\omega$ (note that there are $\omega$ non-zero entries which can each be plus or minus one). Thus if the security is compromised by finding a collision in $H$ this should require $O(2^{\omega/2}\sqrt{\binom n\omega})$ evaluations of $H$ to find. For any given security level, it is possible to find appropriate values of $n$ and $\omega$ for which the required work is greater than the security level.
The easiest way to practically construct such an $H$ is to adapt a regular $h$-bit hash function that is believed to behave like a random oracle. Use this to generate a uniform value between 0 and $V:=2^\omega\binom n\omega$ (e.g. by treating the hash output as a $h$-bit integer; if this value is less than $2^h\mod V$, append a 1 to the input and iterate, else reduce the value modulo $V$). Now split the value $v$ into two values $c:=v/2^\omega$ and $b:=v\mod {2^\omega}$ (note that $b$ and $c$ will be independent and uniformly distributed modulo $2^\omega$ and $\binom n\omega$ respectively). Now use $c$ and the method of this answer to choose a set of $\omega$ coefficients that will be non-zero and use the bits of $b$ to select between the coefficients plus and minus 1.
