$\let\eps\varepsilon$No. I will present a graph whose realization necessarily contains a pair of vertices at distance $1/2$. THis cannot happen if the vertices are algebraic integers.

Firstly, we note that we may force a graph to contain a given piece of triangular lattice. It will be clear after we understand how to enforce the two opposite vertices of a $60^\circ-120^\circ$ unit-sided rhombus to be distinct. This is made by augmenting this rhombus to a Mosers' spindle: the vertices of its realization cannot coincide.

Thus we may enforce two vertices $A$ and $B$ to have any distance realized in the triangular lattice.

Now consider five points $A$, $B$, $C$, $D$, $E$ such that $AB=AC=AD=2$, $BC=DC=BE=DE=1$. If $B\neq D$ and $C\neq E$ then we have $CE=1/2$ as required. Thus it remains to enforce these relations.

To enforce $B\neq D$ it suffices to introduce a point $X$ with $BX=2$ and $DX=1$. To enforce $C\neq E$ it suffices to introduce $Y$ with $CY=2$ and $EY=\sqrt 3$.

**REMARK (expanded).** Maehara showed that any algebraic number can be realized as a distance between two vertices of a rigid unit-distance framework. This inspired the answer; I just needed to make the graph "more rigid". A similar reinforcing may be applied to any rigid framework to make it "absolutely rigid".

To perform this, for every two points $A$, $B$ at distance $d$ in the fixed rigid realization, one needs to ensure that $d-\eps<AB<d+\eps$ in each realization.

To ensure $AB>d-\eps$, find in the triangular lattice two distances $\ell_1$ and $\ell_2$ with $d-\eps <\ell_1-\ell_2<d$ and introduce a point with $AX=\ell_1$, $BX=\ell_2$.

Now, similarly to the construction above, one may realize the distance $1/n$ for every $n\in\mathbb N$. Finally, connecting $A$ and $B$ by an appropriate chain of segments of length $1/n$ one ensures that $AB<d+\eps$.

finitegraphs. $\endgroup$ – Robert Israel Jan 26 '15 at 1:59