6. A.

According to the problem, we can set $a, b, c, d$ as $b-m, b, b+m, \frac{(b+m)^{2}}{b}$ (where $m$ is a positive even number, and $m<b$).
From $d-a=90$, we get $\frac{(b+m)^{2}}{b}-(b-m)=90$, which simplifies to
$$m^{2}+3 b m-90 b=0 \text {. }$$

Since $a, b, c, d$ are even numbers, and $0<a<b<c<d$, we know that $m$ is a multiple of 6, and $m<30$.
$$\begin{array}{l} \text { Let } m=6 k, \text { substituting into equation (1) gives } \\ 36 k^{2}+18 b k-90 b=0 . \end{array}$$

Solving for $b$ gives $b=\frac{2 k^{2}}{5-k}$.
Substituting $k=1,2,3,4$ one by one, and combining with the given conditions, we find that only $k=4, b=32$ is valid. Thus, $m=24$.
Therefore, $a, b, c, d$ are $8, 32, 56, 98$, respectively, so
$$a+b+c+d=194 .$$