Adam has shown that Vopěnka's principle implies that there is no such $\phi$ as in the question. Let me prove this conclusion under a weaker hypothesis, a large cardinal assumption weaker than VP. Specifically, I claim that if there is a stationary proper class of Woodin cardinals, then there is no such $\phi$ as in the question.

The assumption that there is a stationary proper class of Woodin cardinals is strictly weaker than a supercompact cardinal, which is strictly less than an extendible cardinal, which is strictly less than Vopěnka's principle (assuming consistency).

Let's begin. Suppose that the Woodin cardinals form a stationary proper class, and suppose that $\phi:\text{Ord}\to V$ is a function with the stated properties. Since there is a closed unbounded set of $\delta$ with $\phi''\delta\subset V_\delta$, there must be such a $\delta$ that is a Woodin cardinal. Consider $\phi\upharpoonright\delta:\delta\to V_\delta$. Since $\delta$ is Woodin, there is $\kappa<\delta$ such that there is an elementary embedding $j:V\to M$ with critical point $\kappa$ and for which $j(\phi)$ and $\phi$ agree up to and including $\kappa$. In particular, $\phi(\kappa)=j(\phi)(\kappa)$.

Consider now the elements of $\phi(\kappa)$. If we have a natural number $n\in\phi(\kappa)$, then by elementarity, we have $n=j(n)\in j(\phi(\kappa))=j(\phi)(j(\kappa))$. And if there is $x\in \phi(\kappa)$ that is not a natural number, then $x=\phi(\beta)$ for some $\beta<\kappa$. By our assumption on $j$, we know $j(\phi)(\beta)=\phi(beta)=x$. Since $\phi(\beta)\in\phi(\kappa)$, it follows by elementarity that $j(\phi)(j(\beta))=j(\phi)(\beta)\in j(\phi)(j(\kappa))$, which means $x\in j(\phi)(j(\kappa))$. Thus, we've shown $\phi(\kappa)\subset j(\phi)(j(\kappa))$. But since $j(\phi)(\kappa)=\phi(\kappa)$, this means $j(\phi)(\kappa)\subset j(\phi)(j(\kappa))$. So $j(\phi)$ sometimes has inclusion comparable elements, and by elementarity there must be $\alpha<\kappa$ with $\phi(\alpha)\subset\phi(\kappa)$, contradicting one of the properties of $\phi$. QED

Adam has shown that Vopěnka's principle implies that there is no such $\phi$ as in the question. Let me prove this conclusion under a weaker hypothesis, a large cardinal assumption weaker than VP. Specifically, I claim that if there is a stationary proper class of Woodin cardinals, then there is no such $\phi$ as in the question.

The assumption that there is a stationary proper class of Woodin cardinals is strictly weaker in consistency strength than a supercompact cardinal, which is strictly less than an extendible cardinal, which is strictly less than Vopěnka's principle (assuming consistency).

Let's begin. Suppose that the Woodin cardinals form a stationary proper class, and suppose that $\phi:\text{Ord}\to V$ is a function with the stated properties. Since there is a closed unbounded set of $\delta$ with $\phi''\delta\subset V_\delta$, there must be such a $\delta$ that is a Woodin cardinal. Consider $\phi\upharpoonright\delta:\delta\to V_\delta$. Since $\delta$ is Woodin, there is $\kappa<\delta$ such that there is an elementary embedding $j:V\to M$ with critical point $\kappa$ and for which $j(\phi)$ and $\phi$ agree up to and including $\kappa$. In particular, $\phi(\kappa)=j(\phi)(\kappa)$.

Consider now the elements of $\phi(\kappa)$. If we have a natural number $n\in\phi(\kappa)$, then by elementarity, we have $n=j(n)\in j(\phi(\kappa))=j(\phi)(j(\kappa))$. And if there is $x\in \phi(\kappa)$ that is not a natural number, then $x=\phi(\beta)$ for some $\beta<\kappa$. By our assumption on $j$, we know $j(\phi)(\beta)=\phi(\beta)=x$. Since $\phi(\beta)\in\phi(\kappa)$, it follows by elementarity that $j(\phi)(j(\beta))=j(\phi)(\beta)\in j(\phi)(j(\kappa))$, which means $x\in j(\phi)(j(\kappa))$. Thus, we've shown $\phi(\kappa)\subset j(\phi)(j(\kappa))$. But since $j(\phi)(\kappa)=\phi(\kappa)$, this means $j(\phi)(\kappa)\subset j(\phi)(j(\kappa))$. So $j(\phi)$ sometimes has inclusion comparable elements, and by elementarity there must be $\alpha<\kappa$ with $\phi(\alpha)\subset\phi(\kappa)$, contradicting one of the properties of $\phi$. QED