It is not contractible. Let us associate to each matrix $A\in SL_2(\mathbb R)$ the following vector $v(A)$. Take an orthogonal matrix $O\in SO_2(\mathbb R)$ such that $OA(e_1)$ is proportional to $e_1$ with a positive coefficient. Then set $v(A)=OA(e_2)$. We get a map to the upper half plane: $$V:SL(2,\mathbb R)\to \{y>0\}$$
Note that the image of confromal matrices is the point $(0,1)$, and the image of any component $\cal F$ is the complement to $(0,1)$. So we only need to construct a path in $\cal F$ whose image is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take such a path $\gamma(t)$ an consider the unique path of matrices $A_t\subset \cal F$ such that $A_t(e_2)=\gamma(t)$.