If you allow equality then it is trivial that every graph has a distinguishing famil of properties. Since you are considering the property "$v$ has exactly one neighbor", I think you are including equality...
Indeed, let $G$ be a graph, let $v_1,\dots,v_n$ be its vertices, and let me construct a 'distinguishing property' $\phi$ for $v_1$: put $\phi=(\exists x_2,x_2,\dots,x_n)\Phi(v_1,x_2,\dots,x_n)$ with $\Phi(x_1,\dots,x_n)$ being the formula which says that all its arguments are distinct and that the $i$th and $j$th arguments are $\sim$-related iff the vertices $v_i$ and $v_j$ are connected in the graph $G$.
Using formulas constructed like this, I think you can answer your (main?) question.
Let me give an example: consider the graph alt text http://img39.imageshack.us/img39/8420/graphc.png
Then the formula $\phi_1(v)$ corresponding to the vertex $v_1$ in this construction is:
$$(\exists x_2,x_3,x_4,x_5)(v\neq x_2\land v\neq x_3\land v\neq x_4\land v\neq x_5\land x_2\neq x_3\land x_2\neq x_4\land x_2\neq x_5$$ $$\land x_3\neq x_4\land x_3\neq x_5\land x_4\neq x_5\land v\sim x_2\land v\sim x_3\land v\not{\sim}x_4\land v\not{\sim}x_5 \land x_2\sim v\land x_2\not{\sim}x_3$$ $$\land x_2\sim x_4\land x_2\not{\sim}x_5\land x_3\sim v\land x_3\not{\sim}x_2\land x_3\sim x_4\land x_3\not{\sim}x_5\land x_4\not{\sim}v\land x_4\sim x_2\land x_4\sim x_3$$ $$\land x_4\sim x_5\land x_5\not{\sim}v\land x_5\not{\sim}x_2\land x_5\not{\sim}x_3\land x_5\sim x_4)$$