No need to use Radon-Nikodým here. I'll assume that $\beta$ is totally finite. See the end of the answer why this is necessary.
Suppose that $\beta E = 0$ whenever $E$ is measurable and $\eta E = 0$ but that the desired $\varepsilon$-$\delta$-condition doesn't hold.
Then there is $\varepsilon \gt 0$ such that for all $\delta \gt 0$ there is $E$ such that $\eta(E) \lt \delta$ but $\beta(E) \geq \varepsilon$. For each $n$ choose $E_n$ such that $\eta(E_n) \lt 2^{-n}$ and $\beta(E_n) \geq \varepsilon$.
Define $E = \bigcap_{N \in \mathbb{N}} \bigcup_{n \geq N} E_n$. Then $ 0 \leq \eta E \leq \inf_{N \in \mathbb N} \eta \bigcup_{n \geq N} E_n \leq \inf_{N \in \mathbb{N}} \sum_{n \geq N} 2^{-n} = 0 $ hence $\eta E = 0$. By hypothesis it follows that $\beta E = 0$ as well. On the other hand, assuming $\beta$ is totally finite, we get $ 0 = \beta E = \lim_{N \to \infty} \beta \bigcup_{n \geq N} E_n \geq \varepsilon \gt 0, $ which is absurd.
Note that, as @copper.hat pointed out in the comments, it is necessary to assume that $\beta$ is totally finite. The $\sigma$-finite example $\beta E = \int_{E} \frac{1}{t}\,dt$ on $(0,1)$ shows this: For Lebesgue measure $\lambda$ on $(0,1)$, the absolute continuity condition “$\lambda E = 0$ implies that $\beta E = 0$” holds, while the $\varepsilon$-$\delta$-characterization doesn't. For every $\delta \gt 0$ we have $\beta(0,\delta) = \infty$ while $\lambda(0,\delta) = \delta$.