Fibre preserving maps of Borel constructions

The answer to the question in the edit is no. Take $G=\mathbb Z$, $X$ to be a point and $Y=\mathbb R$ with the action $n\cdot x = x+n$. Then there are no equivariant maps from $X$ to $Y$ but there are many maps from $EG \times_G X = BG = S^1$ to the cylinder $EG \times_G Y$ over $BG$.

In this special case when $X$ is a point, an equivariant map $X \to Y$ is a fixed point, while a map $EG \times_G X \to EG \times_G Y$ is a homotopy fixed point. The fixed points include in the homotopy fixed points, but in general I don't think there is much one can say about the inclusion.

Similar to Granja's answer: let $X$ be a point, and $Y = EG$, and let $f$ be induced by the diagonal on $EG$.