The first question, if there is a non-losing strategy, I have an answer for: Yes.
Since this is a finite two-person perfect information game without chance, at least one player must have a non-losing strategy, guaranteed by Zermelo's theorem (of game theory).
For Tic-Tac-Toe related games, it can be proven that the first player has this non-losing strategy. (If it is a winning strategy depends on whether or not the second player has a non-losing strategy).
The argument goes something like this (Player 1 = $P_1$, Player 2 = $P_2$):
Suppose there is a non-losing strategy $S$ for $P_2$. Then $P_1$ will start the game with a random move $X$, and for whatever $P_2$ will do, follow strategy $S$ (thus $P_1$ takes on the role of being the second player). Since $S$ is a non-losing strategy, $P_1$ will not lose, which means $S$ is a non-losing strategy for $P_1$.
Note that, if strategy $S$ ever calls for making the move $X$ (which was the original random move), $P_1$ may simply do another random move $X_2$ and then keep following $S$ as if $X_2$ had been the original random move. This is further explained in page 12-13 here.
(EDIT: Since the first move $P_1$ affects what move $P_2$ can do (by rule 2) the latter argument may not apply to this game. Anyone?)
I will quote some results and problems from the book Combinatorial Games: Tic-Tac-Toe Theory by József Beck, some of which were also quoted in this answer.
The terms "win" and "draw" refer to the game as ordinarily played, i.e., the first player to complete a line wins. The term "Weak Win" refers to the corresponding Maker-Breaker game, where the first player ("Maker") wins if he completes a line, regardless of whether the second player has previously completed a line; in other words, the second player ("Breaker") can only defend by blocking the first player, he cannot "counterattack" by threatening to make his own line. (Note that ordinary $3\times3$ tic-tac-toe is a Weak Win.) A game is a "Strong Draw" if it is not a Weak Win, i.e., if the second player ("Breaker") can prevent the first player from completing a line.
Theorem 3.1 Ordinary $3^2$ Tic-Tac-Toe is a draw but not a Strong Draw.
Theorem 3.2 The $4^2$ game is a Strong Draw, but not a Pairing Strategy Draw (because the second player cannot force a draw by a single pairing strategy.)
Theorem 3.3 The $n\times n$ Tic-Tac-Toe is a Pairing Strategy Draw for every $n\ge5.$
For a discussion of Oren Patashnik's computer-assisted result that $4^3$ tic-tac-toe is a first player win, Beck refers to Patashnik's paper:
Oren Patashnik, Qubic: $4\times4\times4$ Tic-Tac-Toe, Mathematics Magazine 53 (1980), 202-216.
Not much more is known about multidimensional tic-tac-toe, as can be seen from the open problems:
Open Problem 3.2 Is it true that $5^3$ Tic-Tac-Toe is a draw game? Is it true that $5^4$ Tic-Tac-Toe is a first player win?
The conjecture that "if there is a winning strategy on $a^d$, there is one also on $a^{d'}$ for any $d'\geq d$" is given as an open problem:
Open Problem 5.2 Is it true that, if the $n^d$ Tic-Tac-Toe is a first player win, then the $n^D$ game, where $D\gt d$, is also a win?
Open Problem 5.3. Is it true that, if the $n^d$ game is a draw, then the $(n+1)^d$ game is also a draw?
To see that the intuition "adding more ways to win can't turn a winnable game into a draw game" is wrong, consider the following example of a tic-tac-toe-like game, attributed to Sujith Vijay: The board is the set $V=\{1,2,3,4,5,6,7,8,9\};\ $ the winning sets are $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\},$ $\{2,4,8\},$ $\{2,6,9\}$. As in tic-tac-toe, the two players take turns choosing (previously unchosen) elements of $V;$ the game is won by the first player to succeed in choosing all the elements of a winning set. It can be verified that this is a draw game, but the restriction to the board $\{1,2,3,4,5,6,7\}$ (with winning sets $\{1,2,3\},$ $\{1,2,4\},$ $\{1,2,5\},$ $\{1,3,4\},$ $\{1,5,6\},$ $\{3,5,7\}$) is a first-player win.
Best Answer
Lulu's comment hits the nail on the head. The first player can force a win. Player 1 puts a $\times$ here: $$\matrix{- & - & - \\ - & - & - \\ - & - & \times & -}$$ If the second player does not put a $\Large\circ$ somewhere on the bottom row, then Player 1 plays like so: $$\matrix{? & ? & ? \\ ? & ? & ? \\ - & \times & \times & -}$$ Now, player 1 has two winning moves, which cannot be blocked. Thus, Player 2 needs to play somewhere on the bottom row.
Let's consider the case where player 2 plays like so:
$$\matrix{- & - & - \\ - & - & - \\ \Large\circ & - & \times & -}$$
Then player 1 plays:
$$\matrix{- & - & - \\ - & - & - \\ \Large\circ & \times & \times & -}$$
which forces:
$$\matrix{- & - & - \\ - & - & - \\ \Large\circ & \times & \times & \Large\circ}$$
Then player 1 plays:
$$\matrix{- & - & - \\ - & \times & - \\ \Large\circ & \times & \times & \Large\circ}$$
which gives player 1 two possible winning moves, and player 2 will lose after player 1's next turn.
Otherwise, if player 2 did not respond with that move, then the board looks like this:
$$\matrix{- & - & - \\ - & - & - \\ - & ? & \times & ?}$$
Player 1's next move is $$\matrix{- & - & - \\ - & \times & - \\ - & ? & \times & ?}$$ which forces $$\matrix{\Large\circ & - & - \\ - & \times & - \\ - & ? & \times & ?}$$ At this point, player 1's next move is $$\matrix{\Large\circ & - & - \\ - & \times & \times \\ - & ? & \times & ?}$$ Player 2 has no way to prevent player 1's victory.