If I am not mistaken, this expectation equals the coefficient of $x^{\varepsilon n}$ in $(1-x)^{\alpha n}(1+x)^{(1-\alpha) n}$ divided by the corresponding coefficient in $(1+x)^n$ (which equals, of course, $\binom{n}{\varepsilon n}$). Such coefficient may be represented as integrals over unit circle (and so over $[0,2\pi]$) and their asymptotics may be calculated by some standard machinery. I do not write down more, because I do not understand, what exactly do you want (say, which small constant is more, $\alpha$ or $\varepsilon$? Is $n$ chosen large after these constants are already fixed, or $n$ tends to infinity simultaneously with constants tending to zero?).
Update.
Such statement seems to be true. Call two values log-equivalent, if their logarithms are equivalemt. We choose $x_0:=\varepsilon/(1-\varepsilon)$, then $\binom{n}{\varepsilon n}$ is log-equivalent to $x_0^{-\varepsilon n} (1+x_0)^n$. Then interpret our integral as $(2\pi i)^{-1}\int f(z)dz/z$ for $f(z)=z^{-\varepsilon n} (1-z)^{\alpha n}(1+z)^{(1-\alpha)n}$, and integral is taken over the countour around 0. Choose cntour $|z|=x_0$ and note that for small $\alpha$ we will have $|f(z)|\leq f(x_0)$ (this is less or more clear: $|1+z|\leq 1+x_0$ and for $\alpha$ close to 0 this is most important, we have to be careful with neighborhood of $x_0$, but it is ok too). Then just apply this estimate for estimating integral.