1. Start with the given equation:

$\frac{1}{x+p}+\frac{1}{x+q}=\frac{1}{r}$

Given $p\ne q$ and $r\ne 0$.

2. Combine the fractions on the left:

$\frac{(x+q)+(x+p)}{(x+p)(x+q)}=\frac{1}{r}$

$\frac{2x+p+q}{(x+p)(x+q)}=\frac{1}{r}$

Cross-multiplying,

$r(2x+p+q)=(x+p)(x+q)$

3. Expand the right-hand side:

$r(2x+p+q)=x^2+(p+q)x+pq$

So,

$x^2+(p+q)x+pq-2rx-r(p+q)=0$

$x^2+(p+q-2r)x+\big(pq-r(p+q)\big)=0$

4. Let the roots be equal in magnitude and opposite in sign. Then if one root is $\alpha$, the other is $-\alpha$.

So the sum of roots is $0$.

For a quadratic $x^2+bx+c=0$, sum of roots $=-b$.

Hence,

$-(p+q-2r)=0$

$p+q-2r=0$

$r=\frac{p+q}{2}$

5. The product of roots is

$pq-r(p+q)$

But for roots $\alpha$ and $-\alpha$, product is

$\alpha(-\alpha)=-\alpha^2$

Thus,

$-\alpha^2=pq-r(p+q)$

Using $r=\frac{p+q}{2}$,

$-\alpha^2=pq-\frac{p+q}{2}(p+q)$

$-\alpha^2=pq-\frac{(p+q)^2}{2}$

So,

$\alpha^2=\frac{(p+q)^2}{2}-pq$

$\alpha^2=\frac{p^2+2pq+q^2-2pq}{2}$

$\alpha^2=\frac{p^2+q^2}{2}$

6. The roots are $\alpha$ and $-\alpha$. Therefore, the sum of their squares is

$\alpha^2+(-\alpha)^2=2\alpha^2$

$=2\cdot \frac{p^2+q^2}{2}$

$=p^2+q^2$

7. Hence the correct option is:

$\boxed{\text{B: } p^2+q^2}$