Triangles in rigid Riemann surfaces Edit: We thank Vladimir Matveev for his comment on this post which leeds us to  revise the question as follows:
Assume that $M_{g}$ is a compact Riemann surface  with constant negative cuvature (That is $g>1$).
Assume that $ABC$ is  a  triangle on $M_{g}$. Is there a triangle $A'B'C'$ such that $AB=A'B',\;\;AC=A'C',\;\;BC=B'C'$ but there is no an isometry which carries the first triangle to the second one. (Note that the sides of triangles are geodesics.We can consider both cases minimizing or not minimizing  geodesics)
As  another question: Assume that ABC and A'B'C' are two triangles which satisfies the above edges equality and they are homologues to zero. Does this imply  that their interiors have the same volume?
 A: If by isometry, you mean "isometry of the surface", the answer is NO. Indeed, the isometry group of a hyperbolic surface is finite. So, pick a small triangle $\Delta=ABC,$ and let $\Delta(\theta)$ be $\Delta$ rotated by an angle $\theta$ about some point $x_0.$ By finiteness of the isometry group, there is a finite set of $\theta for which there is an isometry.
For the second, the answer is obviously NO. Pick your surface in such a way that there are two separating geodesics of the same length (but which bound different pieces of the surface). Since such a geodesic can be thought of as a triangle, you have your answer.
A: Am I right to assume that you wanted to ask weather the two triangles on the same surface are congruent after lifting them to the universal cover (the hyperbolic plane)? Then the answer is again no, in the case when we don't know what kind of subsurface of the surface each triangle bounds (on both sides). For instance one triangle could bound a disk, so it lifts to a standard hyperbolic triangle, but the other could bound a hyperbolic torus with polygonal (triangular) boundary on one side and a surface of even higher genus on the other (the boundary being the triangle). By the way, it doesn't even have to bound anything specific, i.e. it might have a non-zero homology class. As Igor Rivin pointed out, a special case is a simple closed geodesic whose length is equal to the perimeter of a triangle on the surface that lifts to a standard triangle on the hyperbolic plane via the universal covering map. Then we can choose three points on the geodesic in question so that it gets split into three segments, each equal in length to the corresponding edge from the other triangle.  Analogous arguments (more or less similar to Igor Rivin's above) give also an answer to your second question, which is again no. There are even more non-trivial examples where both triangles are strictly "convex". i.e. the angles at their vertices (on one side) are less than $\pi/2$. In the closed geodesic example, draw three very very small geodesic segments of the same length $\varepsilon$ perpendicular to the closed geodesic at the three points we have already fixed. All segments should be on "one side" of the geodesic. Then the other three endpoints would form such a "convex triangle. Then perturb the other, "trivial" triangle so that it has the same corresponding edge-lengths. You can carry our this construction even on a pair of pants without the need to visualize the whole surface. It is much easier. The bottom line is, that homology is not strict enough. Probably you want the two triangles to be freely homotopic to zero on the surface, i.e. they both bound disks? Then the answer in this case should be yes, they are congruent, after lifting them the the hyperbolic plane (please, correct me if I am wrong here).    
Edit: Maybe I am trying to fit the question to the answers given so far, but it seems to me the question is roughly meant to sound something like that: Let $M$ be a compact hyperbolic surface and assume we have three geodesic segments $AB, BC, CA$ on the surface forming one simple triangular piece-wise geodesic curve $\Delta$ and three more $A'B', B'C', C'A'$ forming another such curve $\Delta'$. It is given that $|AB|=|A'B'|, \, |BC|=|B'C'|, \, |CA|=|C'A'|$, where $|AB|$ is the hyperbolic length of the geodesic segment $AB$, etc. 
Is there an isometry $h : N(\Delta) \to N(\Delta')$ between (small enough) tubular neighborhoods $N(\Delta)$ and $N(\Delta')$ of $\Delta$ and $\Delta'$ respectively (collars, if you prefer this terminology), which maps $\Delta$ to $\Delta'$, i.e. $h(\Delta) = \Delta'$? 
The answer is in general no, but in some cases, (for example when both triangular curves can be freely homotoped to points) there is. But there are other, more non-trivial cases, in which such isometry exists. For instance, since $\Delta$ is a simple closed loop on $M$ there exists a unique simple closed geodesic $\delta$ freely homotopic to $\Delta$ (in the non-trivial case). The same is true for $\Delta'$. Denote by $\delta'$ the simiple closed geodesic freely  homotopic to $\Delta'$. Assume for simplicity that $\Delta$ and $\delta$ bound a cylinder embedded in $M$. Assume the same is true for $\Delta'$ and $\delta$. Then, if you take the points $A, B, C$ and draw on the cylinder bounded by $\Delta$ and $\delta$ the shortest geodesic perpendiculars from each of these three points to the geodesic $\delta$, you obtain three geodesic segments of length $h_A, h_B, h_C$ orthogonal to $\delta$. Do the same for $\Delta'$ and $\delta'$. Then $|AB|=|A'B'|, \, |BC|=|B'C'|, \, |CA|=|C'A'|$ and $h_A=h_{A'}, \, h_B=h_{B'}, \, h_C=h_{C'}$ if and only if there are two tubular neighborhoods that are isometric together with $\Delta$ and $\Delta'$. In this case, $|\delta|=|\delta'|$ (this is a necessary condition). To generalize to non embedded cylinders, consider oriented distances to between the vertices of the triangles and the corresponding simple closed geodesics. 
