Standard deviation of filtered noise
Thank's to MattL's comment I found the 'answer' myself... I was mixing up equations (1) and (2).

As MattL. pointed out, it makes sense to take the square root of equation (1), resulting in something like $$\min_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|\sigma_x\le
\sigma_y\le\max_{\theta\in [\theta_1,\theta_2]}\left|H(e^{j\theta})\right|
\sigma_x\tag{1}$$ if all terms are positive.

If one takes equation (2), applies the integral and the same scaling as in (3) the result is $$\frac{1}{2\pi}\int_{-\pi}^{\pi}S_y(e^{j\theta})d\theta=\frac{1}{2\pi}\int_{-\pi}^{\pi}S_x(e^{j\theta}) \left|H(e^{j\theta})\right|^2 d\theta \tag{2}$$

The left hand side equals the variance of y, so taking the square root gives the standard deviation. However the right hand side is different, as $\left|H(e^{j\theta})\right|$ is a function of frequency.

Hence in general  $\sigma_y\ne |H(e^{j\theta})|\sigma_x$.