Suppose $X, Y$, and $Z$ are topological spaces. If necessary, you may assume that they are nice (manifolds). I am looking for the following result to be true even without this assumption however, so any comments to that effect would also be appreciated.
Let $p:Z\to X$ be a covering map. Let $f:Y\to X$ be a continuous map. Let $x_0\in X$, $y_0\in Y$, $z_0\in Z$ be such that $f(y_0)=x_0=p(z_0)$. Assume that for each loop $\gamma$ at $y_0$ in $Y$, there is a loop $\tilde{\gamma}$ at $z_0$ in $Z$ such that $f\circ \gamma$ is homotopic to $p\circ \tilde{\gamma}$ in $X$ with endpoints both fixed at $x_0$ by the homotopy.
Prove that $f$ lifts to a continuous map $\tilde{f}:Y\to Z$ with $\tilde{f}(y_0)=z_0$.
Overall, it is entirely unclear to me why such a lift should even exist. The standard results concerning relations between fundamental groups are not something that I wish to employ since there are no restrictions on the spaces, nor do I think this would even be helpful. Any comments at all would be appreciated.
You say you do not want to use fundamental groups, but this does not make sense.
In terms of fundamental groups, your condition means that $f_*(\pi_1(Y,y_0)) \subset p_*(\pi_1(Z,z_0))$. It is a well-known theorem that for $Y$ path-connected and locally path-connected this condition is equivalent to the existence of a lift of $f$. See for example Proposition 1.33 in Hatcher's "Algebraic Topology". Hatcher also gives an example showing that the local path-connectedness assumption cannot be omitted.
You can of course copy Hatcher's (fairly short) proof without mentioning fundamental groups, but what would be the benefit?