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The answer to the OP's question in his para. 5 is ``yes'', I think. Even if $X/\mathbb Q$ has no $\mathbb Q$-point there is still an Albanese torsor $T$ (Lang, p. 45, para. 3), universal with respect to morphisms from $X$ to abelian torsors, and then $H^1(X)$ is isomorphic to $H^1(T)$. Define $A=Aut_T^0$, so that $T$ is a torsor under $A$. The isomorphism $A\times T\to T\times T: (a,t)\mapsto (a(t),t)$ gives $H^1(A)\times H^1(T)\cong H^1(T)\times H^1(T)$. Then divide both sides by the copy of $H^1(T)$ coming from the second projection to get $H^1(A)\cong H^1(T)$, so $A$ is the abelian variety you want: $H^1(X)$ is isomorphic to $H^1(A)$.