The first thing to to with a definition like that is test it against some familiar examples. The sequence $\langle 2^{-n} : n\in\mathbb{N}\rangle$ converges to $0$ in $\mathbb{R}$ and it would seem desirable then to have $0$ be a strong ultralimit as well.
That is not going to work: the open sets that contain $0$ give us the cofinite filter in the sequence, hence you do not get any strong ultralimits this way. But the $0$ is an ultralimit (for every free ultrafilter on $\mathbb{N}$) of this sequence. 
So the answer to the first question is no.

The second question has been answered in the comments: If $a$ is a strong $\mathcal{U}$-limit of $\langle x_i:i\in I\rangle$ in a $T_1$-space then the `sequence' must be constant on a member of $\mathcal{U}$.
This implies that your kind of compactness is equivalent to finiteness in the class of $T_1$-spaces. That finite discrete spaces have the compactness property is clear; on the other hand the comments show that no injective sequence can have a strong ultralimit along a free ultrafilter.