As Peter points out, stable homotopy groups are usually defined using the reduced suspension, which requires $X$ and $Y$ to be based. Let's talk about both situations.

As in the comment above, let $Y$ be the subspace $\{1,1/2,1/3,1/4,\ldots,0\}$ of $\mathbb{R}$, and let $X$ be $\mathbb{N}$ with the discrete topology. There is a weak equivalence $X \to Y$.

Let's take 0 to be the basepoint and start taking suspensions. This map becomes a map from a countable wedge of spheres to a higher-dimensional Hawaiian earring that features in a famous paper of Barratt-Milnor (you can prove that this is homeomorphic by the standard "bijection from compact to Hausdorff" argument).

The paper "Homotopy and homology groups of the $n$-dimensional Hawaiian earring" by Eda and Kawamura essentially shows that for $n > 1$, the map $X \to Y$, on $\pi_n$, becomes the embedding $\oplus \mathbb{Z} \to \prod \mathbb{Z}$ from a countable direct sum to a countable product. Therefore, the map on stable homotopy groups is not an isomorphism. (This would be some very small manifestation of Tom Goodwillie's result from the comments.)

However, you can also define stable homotopy groups using iterated unreduced suspension (the groups only become well-defined after a couple of suspensions due to basepoint issues). The unreduced suspension is more homotopically well-behaved, and in particular preserves weak equivalence because it only collapses along cofibrations.

reducedsuspension, this map becomes a map from a wedge of circles to the Hawaiian earring, and so is not a $\pi_1$-isomorphism. However, this depends on using reduced suspension with a particular basepoint, and I'm not sure what the stable $\pi_0$ is. (Do the groups even stabilize if the space is not well-pointed?) – Tyler Lawson Nov 15 '13 at 13:52