Concentration bound for sum of dependent geometric random variable?
No. It is not true. If the uniformly picked edge $u$ have more neighbors, the probability of edges which have $u$ is more than others! Hence, it's not uniform. Moreover, the number of neighbors of each node is changing over time, and this can be more unbiased over the time. Therefore, your selections are not independent of each other.

A solution to uniformly removing an edge can be indexing all edges from $1$ to $|E|$ (using an adjacency list for example) and then pick a random number from $1$ to $|E|$ as your uniformly picking edge in your graph. And do it for the new graph after removing the randomly picked edge.

Hint: As uniform version will remove an edge each time, the probability in this version can be an upper-bound for the $\mathbb{P}(|E[T]-T|)$ in your solution.