No. The picture over a general base is this: let $A_0\to S_0$ be an ordinary abelian variety over a characteristic $p$ scheme $S_0$, and let $S_n$ ($n\geq 0$) be compatible flat liftings of $S_0$ over $\mathbf{Z}/p^{n+1}$. Let $F\colon S_0\to S_0$ be the Frobenius. Then the pull-back $(F^n)^* A_0$ has a canonical lifting $A_{n,\rm can}$ to $S_n$. 

Thus if $S_0$ is perfect (in which case automatically $S_n = W_{n+1}(S_0)$), then we can undo the Frobenius twist $(F^n)^*$ above, and we obtain a compatible system of lifts $A_n = ((F^{-n})^* A_0)_{n, \rm can}$ to $S_n$, i.e. a formal abelian variety over the formal scheme $S_\infty = \varinjlim_n\, S_n$.

If you prefer, you can also say that $A_0$ has a canonical lifting $\tilde A_n$ to $W_{n+1}(S_0)$ for all $n$ (this is an equivalent point of view taken in a paper by Borger and Guerney). If $S_n$ is as above, there is a canonical map $\phi_n \colon S_n\to W_{n+1}(S_0)$ such that $\phi_n\circ i = j\circ F^n$ where $i\colon S_0\hookrightarrow S_n$ and $j\colon S_0\to W_{n+1}(S_0)$ are the canonical closed immersions. Then $\phi_n^* \tilde A_n \simeq A_{n, \rm can}$ as abelian schemes over $S_n$. 

Unfortunately, in your case $S_0 = \operatorname{Spec} k[\varepsilon]/(\varepsilon^2)$ the Frobenius factors through $\operatorname{Spec} k$, so the above gives you nothing. But it does give you something e.g. for $S_0 = \operatorname{Spec} k[\![t]\!]$.

If I have time, I will later come back and explain better why in your case there is no preferred lift.