Let $f:\mathbb{R}^n\to \mathbb{R}$ be a signed distance function for the set $\mathbb{R}^n \setminus \Omega$, meaning: $$f(p) = \begin{cases} d(p, \partial\Omega), & p \in \Omega, \\ -d(p,\partial\Omega), & p \not\in \Omega,\end{cases}$$ where $d(p, A) = \inf_{q\in A} \|p-q\|_2$ is the shortest euclidean distance between $p$ and $A$. In other words the surface $\partial\Omega$ is given as the set of zeroes of $f$: $$\partial\Omega = f^{-1}(0) = \{p\in \mathbb{R}^3\,:\,f(p)=0\}.$$
Now consider a signed distance bound function $g$ for $f$. This means that $g^{-1}(0) = \partial\Omega = f^{-1}(0)$, $g$ is continuous, and $|g|$ is upper bounded by $|f|$: $|g(p)|\leq |f(p)|$.
If you assume that $f$ is continuously differentiable almost everywhere on some open set covering $\partial\Omega$ then you can unambiguously define the normal as $n(p) = \nabla f(p)$ for $p\in\partial\Omega$ almost everywhere (note that $\|\nabla f(p)\|_2=1$ so the normal is unit length). For what you asked to work one needs that $g$ is differentiable almost everywhere on $\partial\Omega$, and moreover $\nabla g(p) = \lambda \nabla f(p)$ for $p\in\partial\Omega$ and $\lambda>0$, i.e. it is a first order approximation of $f$ around $\partial\Omega$ up to a positive constant $\lambda >0$: $$g(p+v) = g(p) + v\cdot \nabla g(p) + o(\|v\|^2) = v\cdot (\lambda\nabla f(p)) + o(\|v\|^2),\quad p\in\partial\Omega.$$
Now consider $g(p) = -f(p)$. The function $g$ clearly satisfies all of the conditions for a signed distance bound but has a normal that points in the opposite direction. That's one possible fail case. A condition to disallow this would be to require that $0\leq g(p) \leq f(p)$ for $p\in\Omega$ and $0\leq -g(p) \leq -f(p)$ for $p\not\in\Omega$, instead of $|g(p)|\leq |f(p)|$ in the definition of signed distance bound function. Then it cannot happen that $\nabla g(p)$ is outside of the hemisphere around $\nabla f(p)$ as otherwise we could move along it and get that $g(p)$ is negative where $f(p)$ is positive (i.e. in $\Omega$) but then $g(p) <0$ which violates $g(p)\geq 0$ for $p\in\Omega$. So this case is taken care of by a small modification of the definition in Hart's paper.
Now let $\nabla g(p)\ne 0$ and $\nabla g(p) \not \parallel \nabla f(p)$. Consider the vector $w = \nabla g(p) - \bigl(\nabla f(p) \cdot \nabla g(p)\bigr) \nabla f(p)$ which is perpendicular to $\nabla f(p)$ but has a positive dot product with $\nabla g(p)$. The function $\psi(h) = g(p+hw)$ grows faster than $\phi(h) = f(p+hw)$ with $h$ increasing (at least for $h$ sufficiently close to $0$), which means that we can make $g(q) > f(q)$ for $q=p+hw$, $h$ small enough and $p\in\partial\Omega$ since: $$g(p+hw) \approx g(p) + hw \cdot \nabla g(p)> 0 = f(p)+h w \cdot \nabla f(p) \approx f(p+hw).$$ This means that allowing non-parallel and non-zero $\nabla g(p)$ for $p\in\partial\Omega$ would also violate the signed distance bound conditions.
Finally, if $\nabla g(p) = 0$ for every point $p\in\partial\Omega$ (or at least for a subset that is not of measure zero) we have a problem. This can occur in practice for example for a half-space SDF: $f(x,y,z) = z$ and the following SDBF: $$g(x,y,z) = \begin{cases} z, & |z|\geq 1, \\ z^3, & |z|<1.\end{cases}$$ So we additionally need to assume that the SDBF function $g$ is chosen such that $\nabla g(p) \ne 0$ almost everywhere on $\partial\Omega$.
TLDR: You can use the normalized gradient $\frac{\nabla g(p)}{\|\nabla g(p)\|_2}$ of a signed distance bound function $g$ (whose derivative is non-zero) as $\pm$ the normal $n(p)$ at a surface point $p\in\partial\Omega$ since the signed distance bound conditions guarantee that it is parallel to the gradient $\nabla f(p) \equiv n(p)$ of the corresponding signed distance function $f$. Replacing the signed distance bound condition $|g(p)|\leq |f(p)|$ from Hart's paper with $0\leq g(p)\leq f(p)$ for $p\in\Omega$ and $0\leq -g(p)\leq -f(p)$ for $p\not\in\Omega$ allows you to dispense away with the $\pm$ (i.e. you get the correctly facing normal).
Practical considerations: If $g$ deviates strongly from $f$ near the boundary then the numerical approximation of the gradient may require smaller step sizes. But as already mentioned in my comments, getting the step sizes right is a difficult problem even when using the sdf $f$ and not a signed distance bound function (sdbf) $g$.
