A number is irrational if it cannot be written as the quotient of two integers; this is a notion independent of the base in consideration. A rational number have a finite or repeating expansion in any base, and an irrational number will have a non-repeating infinite expansion in any base.
For example, $\frac{1}{3}$ is rational, and we write it $0.333 \ldots_{10}$ in decimal expansion, but in base $3$, for example, we can write it as $0.1_3$, and in base $6$ as $.2_6$. The rational number $\frac{1}{3}$ will have a finite or repeating expansion in any base, but never a non-repeating infinite expansion. The number $\pi$ is irrational, and in base $10$ we write it as $3.141592\ldots_{10}$, but in base $2$ it is $11.001001000\ldots_2$; in both bases the expansion will be infinite and non-repeating.
To answer your question: no, the set of terminating or repeating numbers doesn't depend on the base considered.
We fix a base $b$ ($b \in \Bbb N_{\geq 2}$). If we have a number $x$ with a finite expansion in base $b$, then $x = a_ka_{k-1}\ldots a_0.b_1\ldots b_n$ in base $b$. Multiplying by $b^n$ we get $b^n\cdot x = a_ka_{k-1}\ldots a_0b_1\ldots b_n$, and so $\displaystyle x = \frac{a_ka_{k-1}\ldots a_0b_1\ldots b_n}{b^n}$, a ratio of two integers. This shows $x \text{ has a finite expansion in base }b \Rightarrow x \text{ is rational}$
Now consider $x$ having a repeating infinite expansion in base $b$:
$\begin{align*}x &= a_ka_{k-1}\ldots a_0.b_1\ldots b_n c_1 c_2\ldots c_rc_1c_2 \ldots c_r\ldots \qquad \text{(the repeating part is } c_1\ldots c_r \text{)}\\ &\therefore b^n\cdot x = a_ka_{k-1}\ldots a_0b_1\ldots b_n.c_1 c_2\ldots c_rc_1c_2 \ldots c_rc_1c_2 \ldots c_r\ldots \\ &\therefore b^n\cdot x - a_ka_{k-1}\ldots a_0b_1\ldots b_n = 0.c_1 c_2\ldots c_rc_1c_2 \ldots c_rc_1c_2 \ldots c_r\ldots \\ &= \sum_{i=1}^\infty c_1 c_2\ldots c_r\cdot b^{-ri} = c_1 c_2\ldots c_r \sum_{i=1}^\infty b^{-ri}\end{align*}$
You might recognize this sum as a geometric series:
$\displaystyle \sum_{i=1}^\infty b^{-ri} = \sum_{i=1}^\infty (b^{-r})^i = \sum_{i=0}^\infty (b^{-r})^i -1= \frac{1}{1-b^{-r}} - 1 = \frac{1 - 1 + b^{-r}}{1-b^{-r}} = \frac{1}{b^r-1}$
Then we have:
$\begin{align}b^n \cdot x &- a_ka_{k-1}\ldots a_0b_1\ldots b_n = c_1 c_2\ldots c_r \cdot \frac{1}{b^r-1} \\ \therefore &x = \frac{c_1 c_2\ldots c_r \cdot \frac{1}{b^r-1} - a_ka_{k-1}\ldots a_0b_1\ldots b_n}{b^n}= \frac{c_1 c_2\ldots c_r - (b^r-1)\cdot a_ka_{k-1}\ldots a_0b_1\ldots b_n}{b^n(b^r-1)} \end{align}$
which is ratio of two integers. This proves
$x \text{ has a repeating expansion in base }b \Rightarrow x \text{ is rational}$
Then, since $x$ can only be rational or irrational we have, for any $b \in \Bbb N_{\geq 2}$:
$x \text{ has a finite or repeating expansion in base }b \Leftrightarrow x \text{ is rational}$