For Hausdoff spaces, the answer is yes, $X$ is simply connected. The answer here uses some continuum theory. The key is to notice that $X$ is uniquely arcwise connected.
First notice that you cannot embed $S^1$ into $X$ because of your required property. This means that $X$ contains no simple closed curves (homeomorphic images of $S^1$). It is well-known that a Hausdorff space $X$ is uniquely arcwise connected if and only if it does not contain a simple closed curve (this might be an exercise in a continuum theory book, e.g. Nadler's book). So $X$ must be uniquely arcwise connected.
Now let $\alpha:(S^1,p)\to (X,x)$ be a loop based at $p$. The image of $S^1$ in a Hausdorff space is a Peano continuum (a path-connected, locally path connected, compact metric space). Moreover $\alpha(S^1)$ must be uniquely arcwise connected since it is a path-connected subspace of a uniquely arcwise connected space. But any uniquely arcwise connected Peano continuum is a dendrite and all dendrites are contractible. Since $\alpha$ factors through a contractible space, it is null-homotopic.
For non-Hausdorff spaces, path connected does not even imply arcwise connected so the result could potentially be false.