Let L be a language that fulfills the properties implies by the Pumping lemma for regular languages. Does L necessarily fulfill the corresponding properties of the Pumping lemma for context-free languages as well?
Before giving the answers, some discussion on your direction for the proof:
It is highly unlikely that a proof of this claim would use the containment $REG\subseteq CFL$, since the pumping lemma is not an exact characterization, and if a language satisfies the regular pumping lemma, it doesn't proof that the language is regular.
Thus, the proof will have to go into the "guts" of the lemmas. Now for the proof:
Consider a language $L$ that satisfies the regular pumping lemma. Thus, there exists some $p>0$ such that for every word $w\in L$, if $|w|>p$, then there exists a partition $w=xyz$ such that $|xy|\le p$, $|y|>0$ and for every $i$ it holds that $xy^iz\in L$.
We want to prove that there exists some $q>0$ such that for every word $w\in L$, if $|w|>q$, then there exists a partition $w=uvstr$ such that $|vst|\le q$, $|vt|>0$ and for every $i$ it holds that $uv^ist^ir\in L$.
Seeing things stated as above, the rest is just "pattern matching": Let $q=p$, and consider a word $w\in L$ such that $|w|>q$, then there exists a partition $w=xyz$ as above. Let $u=x,v=y,s=\epsilon,t=\epsilon$, and $r=z$. Clearly $uvstr=xyz=w$. Also, $|xy|\le p$, so $|vst|=|y|\le p$ and $|y|>0$ so $|vt|=|v|=|y|>0$. Finally, for every $i$ it holds that $$uv^ist^ir=xy^iz\in L$$ and we are done.