Can algebraic numbers be compared using only rational arithmetic?

Yes.

For the sledgehammer approach, note that the set of $(c_0,\dots,c_{\deg(\alpha)-1})$ such that $\sum_{n=0}^{\deg(\alpha)-1} c_n\alpha^n>0$ can be defined in the language of real-closed fields, so by the Tarski-Seidenberg theorem can be expressed by polynomial inequalities.

This is sort of circular (or at least overkill), since the proof of that theorem relies on Sturm theory which already gives such an algorithm more directly using polynomial remainder sequences; more efficient algorithms use subresultants/Sturm-Habicht sequences. Yap's "Fundamental Problems in Algorithmic Algebra" puts it this way:

(Generalized Sturm) Let $A$ dominate $B$ and let $α < β$ so that $A(α)A(β) \neq 0.$ Then $$\mathrm{Var}_{A,B}[α, β] = \sum_{γ,r,s}\mathrm{sign}(A^{(r)}(γ)B^{(s)}(γ))$$ where $γ$ ranges over all roots of $A$ in $[α, β]$ of multiplicity $r ≥ 1$ and $B$ has multiplicity $s$ at $γ,$ and $r + s$ is odd.

Here $\mathrm{Var}$ is the difference in the number of sign variations in a generalized Sturm sequence defined by $A$ and $B$ when evaluated at $\alpha$ and $\beta,$ and "dominate" is means that if $\gamma$ is an $r$-fold root of $A,$ it is at most an $r$-fold root of $B.$

My $\gamma$ will be your $\alpha,$ sorry. Taking $A\in\mathbb Q[X]$ to the minimal polynomial of $\gamma,$ and taking $[\alpha,\beta]$ to be a rational interval isolating $\gamma$ as a root of $A,$ we can determine the sign of a rational combination $B(\gamma)=\sum_{n=0}^{\deg(A)-1} c_n\gamma^n$ by the generalized Sturm theorem with $r=1$ and $s=0.$ The value of $\mathrm{sign}(A^{(1)}(\gamma))$ can be baked into the algorithm.