The statement is true for a $K(\pi,1)$, but not true for other $X$.

Here's a sketch: let $X = K(\pi,1)$. Let $\Omega_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi_j(\Omega_0X) = \pi_{j+1}(X)$ for $j >0$).

Let $L_0 X$ be the space of contractible unbased loops. Then the sequence $$ \Omega_0 X \to L_0 X \to X $$ is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L_0X \to X$ is a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.

The statement is true for a $K(\pi,1)$ but not true for other $X$. Finite dimensionality is not relevant.

Here's a sketch: let $X = K(\pi,1)$ and I will assume $X$ has the homotopy type of a CW complex. Let $\Omega_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi_j(\Omega_0X) = \pi_{j+1}(X)$ for $j >0$).

Let $L_0 X$ be the space of contractible unbased loops. Then the sequence $$ \Omega_0 X \to L_0 X \to X $$ is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L_0X \to X$ is a weak homotopy equivalence. It follows that the section $X\to L_0X$ given by the constant loops is also a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.

The statement is true for a $K(\pi,1)$, but not generally true for other $X$.

Here's one proof. Let me use the notation $B\pi$ for the $K(\pi,1)$. There's a fibration $$ \Omega B\pi \to L B\pi \to B\pi $$ whre $LX$ is the space of unbased loops in a space $X$ and the second map in the sequence is evaluation at the basepoint of the circle. The fiber appearing is the based loop space of $B\pi$. It has the homotopy type of the discrete space given by the underlying set of $\pi$.

If $L_0B\pi \subset LB\pi$ is the subspace of contractible loops, then the evaluation $$ L_0B\pi \to B\pi $$ is also a fibration with fiber $\Omega_0 B\pi$, the space of contractible based loops. It's easy to see $\Omega_0 B\pi \subset \Omega B\pi \simeq \pi$ is the connected component of the constant based loop. It follows from this that $\Omega_0 B\pi$ is contractible.

Hence, the map $L_0 B\pi \to B\pi$ is a weak homotopy equivalence––and therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.

The statement is true for a $K(\pi,1)$, but not true for other $X$.

Here's a sketch: let $X = K(\pi,1)$. Let $\Omega_0 X$ be the space of contractible based loops in $X$. It's easy to show that the homotopy groups of this space are trivial in every degree (in degree 0 by definition and in higher degrees since $X$ is a $K(\pi,1)$ space, using the fact that $\pi_j(\Omega_0X) = \pi_{j+1}(X)$ for $j >0$).

Let $L_0 X$ be the space of contractible unbased loops. Then the sequence $$ \Omega_0 X \to L_0 X \to X $$ is a fibration, where the second map is evaluation at the basepoint of the circle. Since the fiber has trivial homotopy groups, it follows that the map $L_0X \to X$ is a weak homotopy equivalence. It's therefore a homotopy equivalence since the spaces in question have the homotopy type of a CW complex.