Positive reals $p$ and $q$ are such that the graph of $y = x^2 - 2px + q$ does not intersect the $x$-axis. Find $q$ if there is a unique pair of points $A, B$ on the graph with $AB$ parallel to the $x$-axis and $\angle AOB = \frac{\pi}{2}$, where $O$ is the origin.
1. **Rewrite the quadratic equation**: The given quadratic equation is \( y = x^2 - 2px + q \). We can complete the square to rewrite it in vertex form:
   \[
   y = x^2 - 2px + q = (x - p)^2 + q - p^2
   \]
   This shows that the vertex of the parabola is at \( (p, q - p^2) \).

2. **Condition for no intersection with the x-axis**: For the graph of \( y = x^2 - 2px + q \) to not intersect the x-axis, the discriminant must be negative:
   \[
   \Delta = (-2p)^2 - 4 \cdot 1 \cdot q = 4p^2 - 4q < 0 \implies p^2 < q
   \]

3. **Points \( A \) and \( B \) on the graph**: For \( AB \) to be parallel to the x-axis, the y-coordinates of \( A \) and \( B \) must be the same. Let \( A = (p - a, f(p - a)) \) and \( B = (p + a, f(p + a)) \). Then:
   \[
   f(p - a) = (p - a)^2 - 2p(p - a) + q = a^2 - p^2 + q
   \]
   \[
   f(p + a) = (p + a)^2 - 2p(p + a) + q = a^2 - p^2 + q
   \]
   Thus, \( A = (p - a, a^2 - p^2 + q) \) and \( B = (p + a, a^2 - p^2 + q) \).

4. **Orthogonality condition**: The points \( A \) and \( B \) form vectors from the origin \( O \). For \( \angle AOB = \frac{\pi}{2} \), the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) must be orthogonal. The dot product of these vectors must be zero:
   \[
   \overrightarrow{OA} = (p - a, a^2 - p^2 + q), \quad \overrightarrow{OB} = (p + a, a^2 - p^2 + q)
   \]
   \[
   (p - a)(p + a) + (a^2 - p^2 + q)^2 = 0
   \]
   Simplifying the dot product:
   \[
   (p^2 - a^2) + (a^2 - p^2 + q)^2 = 0
   \]
   \[
   (a^2 - p^2 + q)^2 = a^2 - p^2
   \]

5. **Solve for \( q \)**: Let \( b = a^2 - p^2 \). Then the equation becomes:
   \[
   (b + q)^2 = b
   \]
   \[
   b^2 + 2bq + q^2 = b
   \]
   \[
   b^2 + (2q - 1)b + q^2 = 0
   \]
   For there to be a unique solution for \( b \), the discriminant of this quadratic equation must be zero:
   \[
   (2q - 1)^2 - 4q^2 = 0
   \]
   \[
   4q^2 - 4q + 1 - 4q^2 = 0
   \]
   \[
   -4q + 1 = 0
   \]
   \[
   q = \frac{1}{4}
   \]

The final answer is \( \boxed{\frac{1}{4}} \).