Does existence of a parallel section on $E$ imply local existence of a parallel section on $\bigwedge^k E$?

The answer to your question is no.

Fix a manifold $M$ of dimension $m \geq 2$ and let $E$ be a trivial rank two bundle over $M$. Since your question is local, we might as well assume that $M$ is just an open subset of $\mathbb{R}^m$ but this won't simplify anything. Let $(e_1,e_2)$ be a global frame for $E$ over $M$ and choose some non-closed one form $\alpha \in \Omega^1(M)$. Define a connection $\nabla = \nabla^{\alpha}$ on $E$ by the formula

$$ \nabla_X(f^1 e_1 + f^2 e_2) = (Xf^1)e_1 + ((Xf^2) + \alpha(X)f^2)e_2. \tag{1}$$

You can readily check that this indeed defines a connection and it satisfies

$$ \nabla_X(e_1) \equiv 0, \nabla_X(e_2) = \alpha(X) e_2,\\ R_{\nabla}(X,Y)(f^1 e_1 + f^2 e_2) = f^2 \cdot d\alpha(X,Y) \cdot e_2. $$

In particular, $R_{\nabla} \neq 0$ so $E$ doesn't have two independent parallel sections with respect to $\nabla$, only one (up to a constant multiple): $e_1$. Let's check whether $\Lambda^2(E)$ has a non-trivial parallel section. A general section of $\Lambda^2(E)$ has the form $g(e_1 \wedge e_2)$ where $g$ is a smooth function. Then

$$ \nabla_X(g(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X(e_1 \wedge e_2)) = (Xg)(e_1 \wedge e_2) + g(\nabla_X e_1 \wedge e_2 + e_1 \wedge \nabla_X e_2) = (Xg + g\cdot \alpha(X))e_1 \wedge e_2.$$

Hence, $\Lambda^2(E)$ has a non-trivial parallel section if and only if we can find a non-zero $g$ which satisfies the equation

$$ dg + g \alpha = 0. \tag{2}$$

This is a partial differential equation for $g$. Wedging the equation with $\alpha$ and taking into account that $\alpha$ is a one-form, we get $$ dg \wedge \alpha + g \alpha \wedge \alpha = dg \wedge \alpha = 0.$$ Taking the exterior derivative of the equation, we get

$$ 0 = d^2g + dg \wedge \alpha + g d\alpha = g d\alpha. $$

Since we are looking for non-zero $g$, we see that a necessary condition for the equation to have a solution is $d\alpha = 0$ which doesn't hold since we assumed $\alpha$ is not closed.

In fact, for the family of connections $\nabla^{\alpha}$ we see that $\nabla^{\alpha}$ has two linearly independent parallel sections if and only if the induced connection on $\Lambda^2(E)$ has one linearly independent parallel section. The curvature of the induced connection on $\Lambda^2(E)$ can be identified simply with $d\alpha$. Indeed, $$R^{ \bigwedge^2\nabla}(X,Y)(e_1 \wedge e_2)=R^{ \nabla}(X,Y)(e_1) \wedge e_2+e_1 \wedge R^{ \nabla}(X,Y)(e_2)=d\alpha(X,Y)e_1 \wedge e_2.$$

Since $\Lambda^2(E)$ is a line bundle, it admits a non-zero parallel section if and only if its curvature tensor $R^{ \bigwedge^2\nabla}$ vanished identically, i.e. if and only if $\alpha$ is closed.

Also, note that formula $(1)$ implies that $Xg + g\cdot \alpha(X)=0$ if and only if $\nabla_X(g e_2)=0 $. This gives an alternative proof for the fact that equation $(2)$ admits a non-zero solution if and only if $\nabla$ is flat.