Can a basis for $\mathbb{R}$ be Borel?

This is a hard question. If only for the fact that we don't really know any models of $\sf ZF$ in which $\Bbb R$ is not well-ordered and there is a Hamel basis for $\Bbb R$ over $\Bbb Q$.

While in this person's opinion, the existence of a Hamel basis should not be equivalent to the fact that $\Bbb R$ is well-ordered, but I am not sure how to prove something like that.

Well, if you assume a little bit of choice($\mathsf{DC}$), the answer is no, because of the following argument.

So suppose $B$ is a Hamel basis of $\Bbb R$ over $\Bbb Q$ that also happens to be a Borel set. For each $n\geq 1$, and each $\bar q=(q_1,\ldots,q_n)\in\Bbb Q^n,$ consider the set

$$B_{\bar q}:=q_1B+\cdots+q_nB,$$

then $B_{\bar q}$ is an analytic set, as it is the direct image of the Borel set $B^n$ via the continuous function $\Bbb R^n\rightarrow \Bbb R$ given by $(x_1,\ldots,x_n)\mapsto q_1x_1+\cdots+ q_nx_n$. We have $$\Bbb R=\bigcup_{\bar q\in\Bbb Q^{<\omega}}B_{\bar q}.$$

Thus as this is a countable union and analytical sets are Lebesgue measurable, we must have that $\mu(B_{\bar q})>0$ for some $\bar q\in\Bbb Q^{<\omega}$. Because of this result we know $$B_{\bar q\frown\bar q}=B_{\bar q}+B_{\bar q}$$ contains an open interval. We may assume this open interval contains $0$ using a function $\Bbb R\rightarrow\Bbb R$ of the form $x\mapsto x+a$ , so that $$\bigcup_{n\geq 1}nB_{\bar q\frown\bar q}=\Bbb R,$$

which implies that any $x\in\Bbb R$ is of the form $$m_1q_1x_1+\cdots+m_nq_nx_n+m_1'q_1y_1+\cdots+m_n'q_ny_n,$$

for some $x_1,\ldots,x_n,y_1,\ldots,y_n\in B$ and some $m_1,\ldots,m_n,m_1',\ldots,m_n'\in\Bbb N$ and this is a contradiction, as $\Bbb Q$ is not a finitely generated abelian group. Therefore there cannot exist such set $B.$

Under the presence of large cardinals and using the axiom of choice there cannot even exist Hamel basis that are projective.

If there is for instance a supercompact cardinal, there exists an elementary embedding $$j:L(\Bbb R)\rightarrow L(\Bbb R)^{Col(\omega,<\kappa)},$$ whenever $\kappa$ is a supercompact cardinal; this is a theorem you can find in Woodin's article Supercompact cardinals, sets of reals, and weakly homogeneous trees. Such embeddings can be gotten using way weaker assumptions.

Here's why:

We can take $L(\Bbb R)^{Col(\omega,<\kappa)}$ as the Solovay's model as $\kappa$ is inaccesible. Because of this MO answer there is no Hamel basis of $\Bbb R$ as a $\Bbb Q$-vector space in $L(\Bbb R)^{Col(\omega,<\kappa)}$. Hence for any $X\in L(\Bbb R)$ we have that in $L(\Bbb R)$, $X$ is not a Hamel basis of $\Bbb R$ over $\Bbb Q$, using the elementarity of $j$.

But $\Bbb R\subset L(\Bbb R)$ and $\Bbb R\in L(\Bbb R)$, hence $X$ is a Hamel basis of $\Bbb R$ over $\Bbb Q$ in $L(\Bbb R)$ if and only it is in $V$. However all projective sets belong to $L(\Bbb R)$, thus no Hamel basis can be projective.