A "suitably generic" set of Cohen reals without forcing?

A Cohen generic real, over a model $M$ of set theory, is an infinite binary sequence $g$ whose initial segments get eventually inside any dense set $D$ of finite sequences with $D\in M$. (The set $D$ is dense, if it contains extensions of any given finite sequence.)

When one speaks informally about "sufficiently generic" Cohen reals, without forcing, what is meant is that the sequence has initial segments that are elements of a lot of particularly relevant dense sets. Such generic sequences are often constructed by the following procedure: first, one enumerates the "requirements", which correspond to having certain initial segments (where any finite sequence can be extended to one meeting the requirement), and thus correspond to certain dense sets $D_0,D_1,\ldots$. Next, one constructs an infinite sequence by successively meeting the requirements, one after the next. Thus, one constructs a sequence that is generic with respect to those dense sets and meets all the requirements.

In your case, since you aim to have a perfect set of reals, one constructs the tree, whose branches will be the desired perfect set. You build your tree by finite initial segments, such that at a given level, the nodes are extended in such a way so as to meet the next requirement for its paths. In this way, the branches through the tree will satisfy all the requirements. The key idea will be to design the requirements on the collection of finite trees, so that every finite tree can be top-extended to a tree meeting that particular requirement. In your case, the requirements will correspond to the functions and particular finite nodes currently at the top of the tree we are building---we want to extend those nodes and a target node so that the function value is sufficiently determined and different from the value of the target. In this way, we gradually build up the tree, so that its paths have the desired property for each instance of independence. The main point is that you will have only countably many requirements to meet, and thus only countably many dense sets.

This kind of argument is pervasive in computability theory, and he method can be used as Liang Yu suggests to build perfect sets exhibiting many other kinds of independence.

Meanwhile, the claim becomes obvious once one has knowledge of forcing, and it is to this familiarity that Marker was presumably alluding. Specifically, if $c_0,\ldots,c_n,d$ are mutually $V$-generic Cohen reals and $f$ is a Borel function with a Borel code in $V$, then $f(c_0,\ldots,c_n)\in V[\vec c]$ and therefore cannot be equal to $d$. So mutually generic reals have the desired independence property for Borel functions coded in the ground model. Further, adding a single $V$-generic Cohen real creates a perfect set of finitely mutually generic Cohen reals. So the desired perfect set exists in $V[c]$. But the assertion that there is such a perfect set has complexity $\Sigma^1_2$, and thus is absolute to $V$ by the Shoenfield absoluteness theorem. So you had the desired perfect set already without forcing.

This is a general argument method that works for many constructions in computability theory---one simply argues that fully generic Cohen reals (or Sacks reals, or random reals, or reals generic for some other kind of forcing) have the desired property, and then appeals to absoluteness to get such reals in the original universe.

This is essentially a recursion theoretic question. Just check how to use Cohen forcing to construct a perfect set of independent hyperarithmetic degrees. (See higher recursion theory of Sacks) Then relativize the proof to the family $\mathcal{F}$.

Or even simpler, just construct a perfect set of independent Turing degrees. Then relativize it to some oracle in which the hyperjump of every function in $\mathcal{F}$ is recursive.