The answer is yes.
A measure-preserving invertible shift $T: X \to X$ on a probability space $(X,\mu)$ is said to be weakly mixing if $\lim_{N \to \infty} \frac{1}{N} \sum_{n=1}^N |\langle f \circ T^{-n}, g \rangle|^2 = 0$ for all $f,g$ in the Hilbert space $L^2(X)_0$ of mean zero square-integrable functions, and strongly mixing if $\lim_{n \to \infty} |\langle f \circ T^{-n}, g \rangle| = 0$ for all such functions. (The minus sign is not important here, and one can replace $T^{-n}$ by $T^n$ if one wishes.) There are examples of systems that are weakly mixing but not strongly mixing; see for instance this previous MathOverflow post for some examples (indeed in certain technical senses a "generic" shift is of this form). Note that weakly mixing shifts have no eigenfunctions in $L^2(X)_0$ (indeed this is an if and only if, by the discrete version of the RAGE theorem).
Such systems $(X,\mu,T)$ are discrete flows, but they can be converted into continuous flows by the standard device of taking a suspension. Namely, let $\tilde X$ be $X \times {\bf R}/\sim$ where we quotient by the equivalence relation $(x,t) \sim (T^{-n} x, t+n)$ and endow this space with the product measure $\tilde \mu$ of $\mu$ and Lebesgue measure on the unit interval, and the continuous shift $\tilde T^t (x,s) := (x,s+t)$. If one then lets $H \equiv L^2([0,1]; L^2(X)_0)$ be the Hilbert space of functions $f \in L^2(\tilde X)$ that are of mean zero on every time slice $X \times \{t\}$, and lets $U(t): H \to H$ be the Koopman operator $U(t) f(x,s) := f \circ \tilde T^{-t}(x,s) = f(x,s-t)$, one can easily verify that $U(t)$ is a strongly continuous unitary flow (and thus of the form $e^{itA}$ by Stone's theorem) that has no eigenfunctions, but such that $U(t)$ fails to weakly converge to zero (even if we restrict $t$ to the integers, in which case the continuous flow basically collapses back to the discrete flow $T^n$).
UPDATE: Here is another example. Let $\mu$ be an atomless compactly supported probability measure on ${\bf R}$ whose Fourier transform does not decay to zero at infinity (for instance one can take the standard Cantor set measure). If we let $U(t): L^2(\mu) \to L^2(\mu)$ be the modulation flow $U(t) f(x) = e^{itx} f(x)$, then the flow has no eigenfunctions (because of the atomless condition) but $U(t) 1$ does not converge weakly to zero (because $\langle U(t) 1, 1 \rangle$ is basically the Fourier transform of $\mu$).
