Let $X$ and $Y$ be two nonsingular projective varieties defined over the complex numbers. 



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> If $X$ and $Y$ are $\mathbb{A}^1$-homotopy equivalent, then does $X$ satisfying the Hodge conjecture imply that $Y$ satisfies the Hodge conjecture?
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Is the Hodge Conjecture an $\mathbb{A}^1$-homotopy invariant?
Isn't this very easy? If the varieties are $\mathbb{A}^1$-homotopy equivalent, then their Voevodsky motives are isomorphic also (since there is a connecting functor making the obvious diagram commutative). So it remains to note that all the ingredients of your question are "motivic". For this purpose one may recall that Chow motives embed into Voevodsky one and note that both the cycle classes and the Hodge classes are determined by Chow motives of the corresponding varieties. 


