Yes.
One way to prove it is using Fernique's theorem:
Let $\mu$ be a centered Gaussian measure on a separable Banach space $(X, \|\cdot\|)$. Then there exists $\alpha > 0$ such that $$\int_X e^{\alpha \|x\|^2} \,\mu(dx) < \infty.$$
Using Chebyshev's inequality, it follows that for all $r$,
$$\mu(\{x : \|x\| > r\}) \le C e^{-\alpha r^2}$$
where $C := \int_X e^{\alpha \|x\|^2} \,\mu(dx) < \infty$.
Let's replace $L^2(\mathcal{D})$ with an arbitrary separable Banach space $Y$. Now, the law of your Brownian motion $\{W_t : 0 \le t \le 1\}$ is a centered Gaussian measure on the separable Banach space $X = C([0,1]; Y)$ equipped with the sup norm. Fernique's theorem then tells us that
$$\mathbb{P}\left(\sup_{0 \le s \le 1} \|W_s\|_{Y} > r\right) \le C e^{-\alpha r^2}$$ for appropriate constants $C,\alpha$. Using the Brownian scaling, we thus have for any integer $n$, $$\mathbb{P} \left( \frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y > \epsilon \right) = \mathbb{P} \left( \sup_{0 \le s \le 1} \|W_s\|_Y > \epsilon \sqrt{n} \right) \le C e^{-\alpha \epsilon^2 n}.$$
By the Borel-Cantelli lemma, it follows that $$\mathbb{P} \left( \frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y > \epsilon \quad \text{i.o. $n$}\right) = 0$$ and letting $\epsilon \downarrow 0$ along a sequence, $$\frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y \to 0, \quad \text{a.s.}$$
Noting that when $n-1 \le t \le n$, we have $\frac{W_t}{t} \le \frac{1}{n-1} \sup_{0 \le t \le n} \|W_t\|_Y$, it follows that $\frac{\|W_t\|_Y}{t} \to 0$ almost surely. That is $\frac{W_t}{t} \to 0$ in $Y$-norm almost surely.