What is the equivalent of rotation matrix in Euler angle space for Quaternions?
It's not true to say quaternions don't have a rotation matrix. They are both ways of completely describing a rotation in 3D, so they are freely convertible. Each quaternion corresponds to exactly one rotation matrix, and each rotation matrix corresponds to one quaternion and its negative. Quaternions are just more compact and easier to interpolate.

To work out the rotation matrix for a quaternion, you need to take the effect of the quaternion, $q = (w, x, y, z) = (w, \textbf{v}) $ on some point $\textbf{p}$, that is, $(w, \textbf{v}) * (0, \textbf{p}) * (w, \textbf{-v}) $, and express this formula as a matrix operating on $\textbf{p}$.

Multiplying out using the rules of quaternions gives us $(w^2 - \textbf{v.v})\textbf{p} + 2( w\textbf{v}\times \textbf{p} + \textbf{(v.p)v})  $

This can be rearranged into 3 matrix terms - the first part is just a constant value down the diagonal, the second part with the cross product has values everywhere except the diagonal, and the third one is simply $\textbf{v}_m *  \textbf{v}_n $ for rows m and columns n.

Adding these 3 matrices together gives you 

$\begin{bmatrix} 1-2(y^2+z^2) &amp;2(xy-zw) &amp; 2(xz+yw)\\2(xy + zw)&amp;1-2(x^2+z^2) &amp; 2(yz-xw)\\2(xz-yw) &amp; 2(yz + xw) &amp; 1 - 2(x^2+y^2)\end{bmatrix}$

which is the conversion you need.

In your particular use case, though, you are just doing a simple rotate-x through 90 degrees, which simplifies things a lot. Remember a quaternion is $ ( cos(\theta/ 2) , sin(\theta/ 2)\textbf{a})$   where $\textbf{a}$  is your normalized axis of rotation. So the quaternion representing ±90 degrees rotation around x will be $( \frac{1}{\sqrt{2}},±\frac{1}{\sqrt{2}},0,0)$