The limit equals $-\pi\log 2$, in accordance with Henri Cohen's remark above.

For the proof, we combine the formula
$$r_2(k):=\#\{(m,n)\in\mathbb{Z}^2:m^2+n^2=k\}=4\sum_{d\mid k}\chi_4(d)$$
with Dirichlet's hyperbola method. With this notation, the OP's sum equals
$$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=4\sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de}.$$
We shall now use the well-known facts that
$$\sum_{d=1}^\infty\frac{\chi_4(d)}{d}=\frac{\pi}{4}
\qquad\text{and}\qquad
\sum_{e=1}^\infty\frac{(-1)^e}{e}=-\log 2,$$
where both series are alternating. Using this observation,
\begin{align*}
\sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de}
=&\sum_{\substack{d\leq r\\e\leq r^2/d}}\frac{(-1)^{e}\chi_4(d)}{de}
+\sum_{\substack{e\leq r\\d\leq r^2/e}}\frac{(-1)^{e}\chi_4(d)}{de}-
\sum_{d,e\leq r}\frac{(-1)^{e}\chi_4(d)}{de}\\
=&\sum_{d\leq r}\frac{\chi_4(d)}{d}\left(-\log 2+O\left(\frac{d}{r^2}\right)\right)\\
&+\sum_{e\leq r}\frac{(-1)^e}{e}\left(\frac{\pi}{4}+O\left(\frac{e}{r^2}\right)\right)\\
&-\left(\sum_{d\leq r}\frac{\chi_4(d)}{d}\right)
\left(\sum_{e\leq r}\frac{(-1)^e}{e}\right)\\
=&-\log 2\sum_{d\leq r}\frac{\chi_4(d)}{d}+\frac{\pi}{4}\sum_{e\leq r}\frac{(-1)^e}{e}+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\
=&-\frac{\pi}{4}\log 2-\frac{\pi}{4}\log 2+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\
=&-\frac{\pi}{4}\log 2+O\left(r^{-1}\right).
\end{align*}
Therefore, the OP's sum is
$$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=-\pi\log 2+O\left(r^{-1}\right).$$