Let $\{M_i\}$ be a sequence of 2-dimensional orientable closed surfaces of genus $g$ with smooth Riemannian metrics with the Gauss curvature at least $-1$ and diameter at most $D$. By the Gromov compactness theorem, one can choose a subsequence converging in the Gromov-Hausdorff sense to a compact Alexandrov space with curvature at least $-1$ and Hausdorff dimension 0,1,or 2. One can show (see below) that if $g\geq 2$ then the limit space cannot be a point, thus the dimension of the limit space is at least 1 (while for $g=0,1$ it can be 0).
Let us assume that the limit space has dimension 1. Then it is either circle or segment.
Whether these both possibilities (circle and segment) can be obtained in the limit?
ADDED: It is not hard to see that one can get segment for $g=0$ and circle for $g=1$. I suspect (but cannot prove) that for $g\geq 2$ and $g=0$ one cannot get circle in the limit. In fact I do not even know whether in the case $g\geq 2$ a 2-dimensional limit is the only possibility.
UPDATE: Based on the answer by Igor Belegradek, let me summarize the situation. Let $\{M_i\}$ be a sequence of genus $g$ orientable closed surfaces with Riemannian metrics with Gauss curvature at least -1 which converges in the Gromov-Hausdorff sense to an Alexandrov space $X$.
1) If $g=0$ then $X$ is either a point, or a segment, or $X$ is homeomorphic to $S^2$ (by Perelman stability theorem), and all the three cases are possible.
2) If $g=1$ then $X$ is either a point, or a circle $S^1$, or homeomorphic to the 2-torus, and all the three cases are possible.
3) If $g\geq 2$ then $\dim X=2$ and hence $X$ is homeomorphic to an orientable genus $g$ closed surface.
ADDED: Let me add a proof that if $g\geq 2$ then a point cannot be the limiting space. Indeed otherwise we would have $d_i:=diam(M_i)\to 0$. Let us divide the metric of $M_i$ by $d_i$ and denote the new metric space by $N_i$. Then the sectional curvature of $N_i$ is at least $-d_i^2$ and diameter 1.
By the Gauss-Bonnet $$4\pi(1-g)=\int_{N_i}K\geq -d_i^2vol(N_i).$$
By the Bishop inequality $vol(N_i)$ is bounded from above. Hence the right hand side in the above inequality tends to 0. Hence $1-g\geq 0$ which is a contradiction.
Best Answer
As mentioned in the comments, if the limit is a circle, then by the Yamaguchi fibration theorem, $M_i$ fibers over the circle, and hence it is a torus (or Klein bottle in the non-orientable case).
If the limit is a segment, then $M_i$ is $S^2$ for all large $i$. One (somewhat heavy handed) way to see this is to apply Corollary 0.4 of Shioya-Yamaguchi's paper. Indeed, the product of $M_i$ and a unit circle, collapses to the cylinder. Hence in Corollary 0.4 we have $g=0$ and $k=2$. Hence for large $i$ the fundamental group of $M_i\times S^1$ is a free product of $\mathbb Z$ and finitely many finite cyclic groups. Such a group cannot have a center unless all the finite cyclic groups are trivial. The circle factor of $M_i\times S^1$ is central in the fundamental group, so $\pi_1(M_i\times S^1)=\mathbb Z$. Hence $M_i$ is a sphere (for large $i$).
This argument uses orientability of $M_i$ because Shioya-Yamaguchi only deal with orientable manifolds.