The answer is the same as the first part: the bigger deck, because no new information is revealed to you when a bunch of the cards are thrown out of the big deck.
Perhaps it is simpler to see if you think about what is happening when you randomly choose two cards. It's kind of like you are throwing out 50 of the original 52.
It doesn't make a difference if you take a break after throwing out the first 26 cards, then throwing out another 24 to 'choose two.' The distribution is the same as it would be if you were choosing from 52.
expanding to something more formal...
Let the set of cards $C=\{R_1,\dots, R_{26}, B_1,\dots, B_{26}\}$ where reds are $R=\{R_1,\dots,R_{26}\}$ and blacks are $B=\{B_1,\dots,B_{26} \}$. Let $\mathcal{T}_i$ be the set of $i$-card decks for $i\geq 2$: $$\mathcal{T}_i=\{T \subseteq C: \ |T|=i\}.$$
Let $\mathcal{D}_S$ be the set of 2-card draws from $S$, so that $$\mathcal{D}_T=\{D \subset T:\ |D|=2\}.$$
Now we are interested in the space:
$$\Omega_i = \underset{T\in\mathcal{T}_i}{\prod} T \times D_T$$
Everything is chosen randomly, so we know if $T\in \mathcal{T}_i$ then $P(T)=1/{|C| \choose i}.$ Also, if $D\in \mathcal{D}_T$ then $P(D|T)=1/{i \choose 2}.$ It is easy to see that this induces a unique probability measure on the entire space. In particular, $P(T,D)=1/\left({|C| \choose i}{i \choose 2}\right)$ if $D\in \mathcal{D}_T$ and 0 otherwise.
Let $X_i$ be a RV on $\Omega_i$ where if $(t,d)\in \Omega_i$ then $X_i(t,d)=1$ if $d\subseteq R$ or if $d\subseteq B$, and 0 otherwise.
You've already determined that the probability that two guys are the same color choosing from the specific deck of 52, $S=\{R_1,\dots, R_{13},B_1,\dots, B_{13}\}$ is $25/51$.
We are also interested in the probability that two guys are the same color choosing from an arbitrary deck of size 26. Or: $P(X_{26}=1).$ This can be rewritten as:
\begin{array}{rl}
& P\left(\cup_{T\in\mathcal{T}_{26}}\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R \lor D \subseteq B\right. \right\rbrace\right) \\
&= \sum_{T\in \mathcal{T}_{26}}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\lor D \subseteq B\right. \right\rbrace\right) \\
&\overset{\text{(a)}}{=} 2\sum_{T\in \mathcal{T}_{26}}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\
&\overset{\text{(b)}}{=} 2\sum_{T\in \mathcal{T}_{26}: |T\cap R|\geq 2}P\left(\left\lbrace D \subseteq \mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\
&\overset{\text{(c)}}{=} 2\sum_{i=2}^{26}{26 \choose i}{26 \choose 26-i }P\left(\left\lbrace D \subseteq \cup_{T: |T\cap R|=i}\mathcal{D}_T\left| \ D \subseteq R\right. \right\rbrace\right) \\
&\overset{\text{(d)}}{=} 2\sum_{i=2}^{26}{26 \choose i}{26\choose 26-i }{i \choose 2}/\left({52 \choose 26}{26 \choose 2}\right)\\
&\overset{\text{(e)}}{=} 25/51.
\end{array}
(a) is due to symmetry in $R$ and $B$ and their disjointness. The probability we get two blacks is the same as the probability that we get two reds.
(b) is due to our inability to get two reds if there are one or less in the deck we are choosing from. This makes those terms have probability 0 and not contribute to the sum.
(c) Fixing $i,$ we enumerate all possible ways of choosing $i$ reds and filling the rest of our small deck with blacks. ${26 \choose i}{26 \choose 26-i}$
(d) By the way we rewrote the sum in our previous step, we know each deck has ${i}$ reds, so the number of 2-draws of red we could make is ${i \choose 2}.$ Further we know the probability of each of these $D$ by previous investigation of our model.
(e) Rewriting the summation terms, we see that it's the same.
Best Answer
Posting my comment as an answer:
If you first pick $26$ cards out of $52$ and then pick two random cards out of those, it's the same as picking $26$ cards out of $52$ and then taking first two of them because all orderings of $26$ cards are equally likely. Which in turn gives the same result as picking $2$ cards out of $52$ and then not picking $24$ others.