The expected length is infinite (when $\epsilon \lt 1/2$)1/2$.
The reason is that when the slope $\alpha$ is near a rational $p/q$ such that there are infinite trajectories of slope $p/q$ which don't get closer than $\epsilon$ to any lattice point, the length of the trajectory is often long. Specifically, suppose $\alpha$ is very close to $0$. Then the expected length of the trajectory is not roughly $1/\epsilon$, it is proportional to $(1-2\epsilon)/\alpha$.
By rotating the lattice, we can assume that the direction of the ray is between $-\pi/4$ and $\pi/4$, and a uniform distribution on these directions has a bounded distortion $c_1$ from the uniform distribution on slopes between $-1$ and $1$. The expected value of a nonnegative quantity over slopes is a factor of at most $c_1$ off of the expected value over directions.
Suppose the slope is between $1/n$ and $2/n.$ For simplicity, assume the initial point has $x$-coordinate $0$. This adds less than $2$ to the length. For simplicity, in the following I'll assume $\epsilon \lt 1/2\sqrt2$ but that isn't necessary. In order to get within $\epsilon$ of a lattice point, the height of the ray has to be within $c \epsilon$ (with $c_2 \lt \sqrt 2, c_2 \lt 1+2/n$) of an integer when the $x$-coordinate is an integer. For an initial $y$-coordinate $y_0$ between $c \epsilon$ and $1-c \epsilon$, a lower bound for when this can occur is $(1-c_2\epsilon)/(2/n) (1-c_2\epsilon)/(2/n)$. So, as we average over the initial $y$-coordinates, the average length is at least $\frac{(1-2c_2 \epsilon)(1/2 - c_2 \epsilon)}{(2/n)} -2 \gt c_3 n$ - 2.
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Since the intervals $(1/2^m, 2/2^m)$ are disjoint, the expected length is at least $c_1 \sum_{n=2^m} (1/n) (c_3 n - 2) = -2 + c_1 \sum_m c_3 = \infty.$
