I originally gave a wrong candidate for an answer, but in fact no such path can exist (at least, assuming that your topologies are such that $t \mapsto \lambda(p, t)$ is continuous for all paths $p$). Indeed, define $p : [0, 1] \to \operatorname M(2, \mathbb R)$ by
$$
p(t) = \begin{pmatrix} \cos(\pi t) & \sin(\pi t) \\ \sin(\pi t) & -\cos(\pi t) \end{pmatrix}
$$
for all $t \in [0, 1]$. Then $\lambda(p, t) \in \{\pm1\}$ for all $t \in [0, 1]$, but $\lambda(p, 1) = -\lambda(p, 0)$, which is impossible if $t \mapsto \lambda(p, t)$ is continuous.

EDIT: I suspected from your mention of the simple connectedness of $\operatorname{SL}(2, \mathbb C)$, and see definitely from your comment, that you are interested in the non-simple connectedness of $\operatorname{SL}(2, \mathbb R)$. It may be that a particularly simple way to see this is to consider the metaplectic group, which is a non-split double cover of $\operatorname{SL}(2, \mathbb R)$. Notice that your original example can be adapted to give a *polynomial* path in $\operatorname{SL}(2, \mathbb R)$ (not just in $\operatorname M(2, \mathbb R)$, or even just in $\operatorname{GL}(2, \mathbb R)$ as my original path was):
$$
t \mapsto p(t) = \begin{pmatrix}
t(1 - t) & 1 - t^2(6 - 4t) \\
-1 + t^2(6 - 4t) & 4t(1 - t)(3 - 2t)(1 + 2t)
\end{pmatrix}.
$$

Notice that the cover of $\operatorname{SL}(2, \mathbb R)$ implicit in your question, namely
$$
E \mathrel{:=} \{(g, \lambda) : \text{$g \in \operatorname{SL}(2, \mathbb R)$ and $\lambda$ is an eigenvalue of $g$}\},
$$
is a branched cover; it is a double cover only on the regular semisimple set (of elements with distinct eigenvalues). Its restriction to the regular semisimple set is isomorphic (as a cover of the rss set) to the restriction there of the branched cover
$$
\{(g, B) : \text{$g \in \operatorname{SL}(2, \mathbb R)$ and $B$ is a Borel subgroup of $\operatorname{SL}(2, \mathbb C)$ containing $g$}\}
$$
(namely, associate $(g, \lambda)$ and $(g, B)$ if and only if the non-trivial eigenvalue of $\operatorname{Ad}(g)$ on $\operatorname{Lie}(B)$ is $\lambda^2$). However, I think that the restriction is *not* isomorphic to the restriction of the metaplectic group. Indeed, let $\lambda$ and $\mu$ be distinct elements of $\mathbb R$ satisfying $\lambda\mu = 1$. Then any lift to $E$ of the path
$$
t \mapsto \operatorname{Int}\begin{pmatrix}
\cos(\pi t) & \sin(\pi t) \\
-\sin(\pi t) & \sin(\pi t)
\end{pmatrix}\begin{pmatrix}
\lambda & 0 \\
0 & \mu
\end{pmatrix}
$$
projects, *via* $E \to \mathbb C$, to $\{\lambda, \mu\}$, hence is constantly $\lambda$ or constantly $\mu$; but, if I've done my computation properly, then
$$
\operatorname{Int}(w(t))\Bigl(\begin{pmatrix}
\lambda & 0 \\
0 & \mu
\end{pmatrix}, \sqrt\mu\Bigr),
$$
where $w(t) = \Bigl(\begin{pmatrix}
\cos(\pi t) & \sin(\pi t) \\
-\sin(\pi t) & \cos(\pi t)
\end{pmatrix}, \epsilon_t\Bigr)$, with $\epsilon_t(i) = e^{\pi i t/2}$, is a path in the restriction to the rss set of the metaplectic cover that connects the two pre-images of $\begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}$. (I'm using the description of points in the metaplectic group from Wikipedia.)

reallyneed the topology you use on the space of paths. $\endgroup$