The answer is "yes". To see this, let us recall the concept of a pullback. Given a pair of maps $(p : E \to X,f : Y \to X)$, a pullback of $(p,f)$ is given by a pair of maps $(p^* : E^* \to Y,f^*: E^* \to E)$ such that
(1) $fp^* = pf^*$
(2) For any pair of maps $(u : Z \to Y, v : Z \to E)$ such that $fu = pv$ there exists a unique map $w : Y \to E^*$ such that $p^*w = u$ and $f^*w = v$.
This universal property implies that pullbacks are unique up to canonical homeomorphism. Here is an explicit construction:
$$E^* = \{ (e,y) \in E \times Y \mid p(e) = f(y) \}, p^*(e,y) = y, f^*(e,y) = e .$$
If $f$ is the inclusion map of a subspace $Y \subset X$, then we may also take $E^* = p^{-1}(Y) \subset E, p^*(e) = p(e), f^*(e) = e$.
It is well-known that pullbacks of covering projections are covering projections. This means that if $p$ is a covering projection, then so is $p^*$. See Pullback of a covering map.
Lemma : Let $p : E \to X$ be a covering projection and $f_0, f_1 : Y \to X$ be homotopic maps. Then the pullback covering projections $p^*_k : E^*_k \to Y$ along $f_k$ are equivalent which means that there exists a homeomorphism $h : E^*_0 \to E^*_1$ such that $p^*_1h = p^*_0$.
Proof. Let $H : Y \times I \to X$ be a homotopy from $f_0$ to $f_1$ and for $k = 0,1$ let $i_k : Y \times \{ k \} \hookrightarrow Y \times I$ denote inclusion. Form the pullback $(p^* : E^* \to Y \times I, H^* : E^* \to E)$ of $(p,H)$ and the pullbacks $(\pi_k : E^*_k \to Y \times \{ k \}, i^*_k : E^*_k \hookrightarrow E^*)$ of $(p^*,i_k)$, where $E^*_k = (p^*)^{-1}(Y \times \{ k \}) $. Then we may assume that $p^*_k = r_k \pi_k : E^*_k \to Y$, where $r_k : Y \times \{ k \} \to Y, r_k(y,k) = y$.
Now the map $i_0 \pi_0 : E^*_0 \to Y \times I$ lifts to $i^*_0 : E^*_0 \to E^*$. Hence the homotopy
$$G : E^*_0 \times I \to Y \times I, G(e,t) = (p^*_0(e),t)$$
which satisfies $G_0 = i_0 \pi_0$ lifts to a homotopy $G' : E^*_0 \times I \to E^*$. The map $G'_1 : E^*_0 \to E^*$ has the property $G'_1(E^*_0) \subset (p^*)^{-1}(Y \times \{ 1 \}) = i^*_1(E^*_1)$. Therefore we get a map $h : E^*_0 \to E^*_1$ such that $p^*_1h = p^*_0$. By lifting $G' : E^*_1 \times I \to Y \times I, G'(e,t) = (p^*_1(e),1-t)$ we obtain a map $h' : E^*_1 \to E^*_0$ such that $p^*_0h' = p^*_1$. It is easy to see that $h' h = id$ (this follows from unique path lifting).
Now let $U \subset X$ be open and contractible. This means that the inclusion $i : U \to X$ is homotopic to a constant map $c : U \to X, c(y) = x_0$. Hence the covering projection $p : p^{-1}(U) \to U$ is equivalent to the pullback of $p$ along $c$. But the latter is easily seen to be a trivial covering $U \times F \to U$ (recall $E^* = \{ (e,y) \in E \times U \mid p(e) = c(y) = x_0 \} = \{ (e,y) \in E \times U \mid e \in p^{-1}(x_0)) \} , p^*(e,y) = y$). This means that $U$ is evenly covered.
Note that it suffices to assume that $i : U \to X$ is inessential which is weaker than $U$ contractible.
