I was asking myself the same question, but under weaker assumptions and I think I finally found an answer. Unfortunately there is a counterexample satisfying even the strong assumptions above. It was motivated by the answer of VF1 for the question regarding the continuous version of the Borel-Cantelli lemma. Unfortunately his answer didn't satisfy the convergence in probability to $0$, so I took another function and modified the process. Please check if the couterexample is correct.
For the sake of simplicity we consider only the idex set $[3,\infty[$ instead of $\mathbb{R}_+$, but the example can be shifted of course.
Consider the test function (i.e. element of $C^\infty_c(\mathbb{R})$)
$$\phi_b(x)=\begin{cases}\exp(b^2/(x^2-b^2),\qquad&x\in]-b,b[\\
0,\qquad&x\notin]-b,b[
\end{cases}$$ for $b>0$ and the probability space $(\Omega, \sigma, P)=([0,1[,\mathcal{B}([0,1[),\lambda)$, whereas $\lambda$ is the Lebesgue measure. Note: $\forall b>0:\phi_b(0)=e^{-1}$ and $\forall n\in\mathbb{N}\forall \omega\in[0,1[:$
$$\phi_{1/n^2}(t-\omega-n)=\begin{cases}\exp(n^{-4}/((t-\omega-n)^2-n^{-4}),\qquad&t\in]n+\omega-n^{-2},n+\omega+n^{-2}[\\
0,\qquad&t\notin]n+\omega-n^{-2},n+\omega+n^{-2}[.
\end{cases}$$
We define the process $Y_t(\omega):=\sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n).$ Note, that $Y$ is continuous and for fixed $\omega$ there is for $t\geq3$ always at most one of the functions inside the sum unequal to $0$. Obviously $Y$ is not converging to $0$ a.s., since $\forall k\in\mathbb{N}:Y_{\omega+k}(\omega)=e^{-1}$. But it holds for all $\epsilon>0$
\begin{align}
P(Y_t>\epsilon)\leq P(Y_t>0)=& P\left(\omega\in[0,1[: \sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n)>0\right)\\
\leq& 3P\left(\omega\in[0,1[: |t-\lfloor t\rfloor-\omega|\leq\frac{1}{(\lfloor t\rfloor-1)^2} \right)\\
\leq&3\lambda\left([t-\lfloor t\rfloor-\frac{1}{(\lfloor t\rfloor-1)^2},t-\lfloor t\rfloor+\frac{1}{(\lfloor t\rfloor-1)^2}]\right)\\
=&\frac{6}{(\lfloor t\rfloor-1)^2},
\end{align}
so it is converging to $0$ in probability. Here in the second inequality we estimate with $\lfloor t\rfloor-1$ instead of $\lfloor t\rfloor$ or $\lfloor t\rfloor+1$, since for big $\omega$ and small $t-\lfloor t\rfloor$ the $t$ can be "in the bump belonging to the previous integer". Analogous for small $\omega$ and big $t-\lfloor t\rfloor$ with the bump of the following integer. The three possible cases cause the factor 3. Furthermore
$$\int_3^\infty P(Y_t>\epsilon)\mathrm{d}t\leq\int_3^\infty \frac{6}{(\lfloor t\rfloor-1)^2}\mathrm{d}t=6\sum_{n=2}^{\infty}\frac{1}{n^2}<\infty
$$ the continuous analogy of the Borel-Cantelli condition holds and
$$\int_3^\infty E(Y_t)\mathrm{d}t=\int_3^\infty E(Y_t\cdot{1}_{Y_t>0})\mathrm{d}t\leq e^{-1}\int_3^\infty P(Y_t>0)\mathrm{d}t\leq e^{-1}\int_3^\infty \frac{6}{(\lfloor t\rfloor-1)^2}\mathrm{d}t<\infty
$$ condition 1) is satisfied. Also condition 2) seems to me beeing satisfied, but it is a lot of calculation
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}t} E(Y_t)=\frac{\mathrm{d}}{\mathrm{d}t}\int_0^1 \sum_{n=1}^\infty \phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)=&\sum_{n=1}^\infty \int_0^1 \frac{\mathrm{d}}{\mathrm{d}t}\phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)\\
=&\sum_{n=1}^\infty \int_0^1 -\frac{\mathrm{d}}{\mathrm{d}\omega}\phi_{1/n^2}(t-\omega-n) \mathrm{d}\lambda(\omega)\\
=&\sum_{n=1}^\infty \phi_{1/n^2}(t-0-n)-\phi_{1/n^2}(t-1-n)\\
=&\sum_{n=1}^\infty \phi_{1/n^2}(t-n)-\phi_{1/n^2}(t-(n+1)).
\end{align}
To see that this is less or equal to zero we consider
\begin{align}
&\sum_{n=1}^\infty \phi_{1/n^2}(t-n)-\phi_{1/n^2}(t-(n+1))\\
&=\phi_{1}(t-1)-\phi_{1}(t-2)+\phi_{1/4}(t-2)-\phi_{1/4}(t-3)+\phi_{1/9}(t-3)-...\\
&=\underbrace{\phi_{1}(t-1)}_{=0,\text{ für } t\geq2}+\underbrace{\big(\phi_{1/4}(t-2)-\phi_{1}(t-2)\big)}_{\leq0}+\underbrace{\big(\phi_{1/9}(t-3)-\phi_{1/4}(t-3)\big)}_{\leq0}+...
\end{align}
Since we consider only $t\geq3$, $\frac{\mathrm{d}}{\mathrm{d}t} E(Y_t)\leq0$ holds and condition 2) is satisfied.
In summary, we found a nonnegative process $Y$ with a.s. $C^\infty([3,\infty[)$ paths and
- $\int_3^\infty P(|Y_t|>\epsilon)\mathrm{d}t<\infty$,
- $\int_3^\infty E(|Y_t|)\mathrm{d}t<\infty$,
- $[3,\infty[\ni t \mapsto E(|Y_t|) $ is decreasing,
- $Y$ is converging to $0$ in probability,
- $Y$ is not converging to $0$ a.s.
Furthermore, from 2. and 3. we can conclude
- $Y$ is converging to $0$ in $\mathcal{L}^1(P)$.
