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 $(0,1)$. And so each component can be identified with this puncutred half-plane.  Hence it is enough to construct a path in $\cal F$ whose image under $V$ is not contractible in $\{y>0\}\setminus \{(0,1)\}$. This is easy, just take a non-contractible path $\gamma(t)\subset \{y>0\}\setminus \{(0,1)\}$ (that has a non-zero winding number around $(0,1)$),  and consider the unique path of matrices $A_t\subset \cal F$ such that $A_t(e_2)=\gamma(t)$.